Is $\ 7!=5040\ $ the largest highly composite factorial?
Suppose $n \ge 20$. Trivially, $n!$ is then divisible by $16$, so $\dfrac{13}{16}n!$ is an integer smaller than $n!$.
Let the prime factorization of $n!$ be $n! = 2^{e_1}3^{e_2}\cdots11^{e_5}13^{e_6}17^{e_7}\cdots p_r^{e_r}$.
Then the prime factorization of $\dfrac{13}{16}n!$ is $\dfrac{13}{16}n! = 2^{e_1-4}3^{e_2} \cdots11^{e_5}13^{e_6+1}17^{e_6}\cdots p_r^{e_r}$.
Then, the number of divisors of $n!$ is $\sigma_0(n!) = (e_1+1)(e_6+1)\displaystyle\prod_{k \neq 1,6}(e_k+1)$ and the number of divisors of $\dfrac{13}{16}n!$ is $\sigma_0(\dfrac{13}{16}n!) = (e_1-3)(e_6+2)\displaystyle\prod_{k \neq 1,6}(e_k+1)$.
Hence, $\sigma_0(\dfrac{13}{16}n!) > \sigma_0(n!)$ iff $(e_1-3)(e_6+2) > (e_1+1)(e_6+1)$ iff $e_1 > 4e_6+7$.
Using the well known formula for the largest power of a prime dividing a factorial, we have
$e_1 = \displaystyle\sum_{k = 1}^{\left\lfloor \log_2 n\right\rfloor}\left\lfloor \dfrac{n}{2^k} \right\rfloor > \displaystyle\sum_{k = 1}^{\left\lfloor \log_2 n\right\rfloor} \left(\dfrac{n}{2^k} - 1\right) \ge n - \dfrac{n}{2^{\left\lfloor \log_2 n\right\rfloor}} - \left\lfloor \log_2 n\right\rfloor \ge n - 2 - \log_2 n$.
Similarly, $e_6 = \displaystyle\sum_{k = 1}^{\left\lfloor \log_{13} n\right\rfloor}\left\lfloor \dfrac{n}{13^k} \right\rfloor \le \displaystyle\sum_{k = 1}^{\left\lfloor \log_{13} n\right\rfloor}\dfrac{n}{13^k} \le \dfrac{n}{12}$. Hence, $4e_6+7 \le \dfrac{n}{3}+7$.
I'll leave it as an exercise to show that $\dfrac{n}{3}+7 < n - 2 - \log_2 n$ holds for all integers $n \ge 20$.
Therefore, for all integers $n \ge 20$, we have $4e_6+7 \le \dfrac{n}{3}+7 < n - 2 - \log_2 n < e_1$, and thus, $\sigma_0(\dfrac{13}{16}n!) > \sigma_0(n!)$.
So, $n!$ is not highly composite for $n \ge 20$. Now, it remains to check if any factorials between $7!$ and $20!$ are highly composite.
EDIT: From the list of the first 1000 highly composite numbers that OP mentioned, it appears that the $149$-th largest highly composite number is $\approx 1.49 \times 10^{17}$, while $19! \approx 1.22 \times 10^{17}$. None of the numbers $8!, 9!, \ldots, 19!$ appear in that list, so $7!$ is indeed the largest highly composite factorial.
Yes, it is known. If you look at the number of divisors function, a number $n$ with prime factorization $p_1^{a_1}p_2^{a_2}p_3^{a_3}\ldots$ has $\sigma_0(n)=(a_1+1)(a_2+1)(a_3+1)\ldots$ divisors. The factorials have too many small factors. Every time you multiply by a prime you double the number of divisors, but when you increase the power of $2$ from $n$ to $n+1$ you only multiply the number of divisors by $\frac {n+2}{n+1}=1+\frac 1{n+1}$. The greedy algorithm for finding highly composite numbers would look at $\log \sigma_0(n)$. If you multiply $n$ by a prime $p$ that is currently in the factorization of $n$ as $p^a$ you add $\log 1+\frac 1{a+1}$ to the log of the number of factors and you add $\log p$ to the log of $n$, so the benefit/cost ratio of each increase is $\frac {\log 1+\frac 1{a+1}}{\log p}$ Search over the primes, find the best, and that is your next multiplier. When you look at OEIS A002182 not every entry is a multiple of the previous one, but you see there are too many small factors among the factorials. $n!$ has about $\frac n2$ factors of $2$, for example. To really justify this you need to use the prime number theorem to show how large the next prime will be, then show you want to use it before $n!$ will. You can find a finite limit of factorials that are candidates, then check them.
What you are missing is the ratio of exponents. For factorials, by the theorem of Legendre, the exponent of $p$ is proportional to $1/p.$ I suppose it is slightly more accurate to say proportional to $1 /(p-1).$
The superior highly composite numbers have exponents for each prime proportional to $1/ \log p.$ Guy Robin made a method for interpolating between consecutive SHC numbers to construct all highly composite numbers between. The point is that highly composite numbers are, in the long run, similar to the least common multiple of the numbers from $1$ to some $m.$
It might be a good deal of work to show that $5040$ is the last factorial that works, but it is clear that there are only finitely many such, and explicit bounds could be constructed. From the output below, you can see how much smaller an SHC number can be, yet have more divisors than the factorial under consideration. The logarithms are base ten.
jagy@phobeusjunior:~$ ./factorial
2! 2 = 2 div 2 log 0.30103
3! 6 = 2 3 div 4 log 0.778151
4! 24 = 2^3 3 div 8 log 1.38021
5! 120 = 2^3 3 5 div 16 log 2.07918
6! 720 = 2^4 3^2 5 div 30 log 2.85733
7! 5040 = 2^4 3^2 5 7 div 60 log 3.70243
8! 40320 = 2^7 3^2 5 7 div 96 log 4.60552
9! 362880 = 2^7 3^4 5 7 div 160 log 5.55976
10! 3628800 = 2^8 3^4 5^2 7 div 270 log 6.55976
11! 39916800 = 2^8 3^4 5^2 7 11 div 540 log 7.60116
12! 479001600 = 2^10 3^5 5^2 7 11 div 792 log 8.68034
13! 6227020800 = 2^10 3^5 5^2 7 11 13 div 1584 log 9.79428
14! 87178291200 = 2^11 3^5 5^2 7^2 11 13 div 2592 log 10.9404
15! 1307674368000 = 2^11 3^6 5^3 7^2 11 13 div 4032 log 12.1165
16! 20922789888000 = 2^15 3^6 5^3 7^2 11 13 div 5376 log 13.3206
17! 355687428096000 = 2^15 3^6 5^3 7^2 11 13 17 div 10752 log 14.5511
18! 6402373705728000 = 2^16 3^8 5^3 7^2 11 13 17 div 14688 log 15.8063
19! 121645100408832000 = 2^16 3^8 5^3 7^2 11 13 17 19 div 29376 log 17.0851
20! 2432902008176640000 = 2^18 3^8 5^4 7^2 11 13 17 19 div 41040 log 18.3861
21! 51090942171709440000 = 2^18 3^9 5^4 7^3 11 13 17 19 div 60800 log 19.7083
22! 1124000727777607680000 = 2^19 3^9 5^4 7^3 11^2 13 17 19 div 96000 log 21.0508
23! 25852016738884976640000 = 2^19 3^9 5^4 7^3 11^2 13 17 19 23 div 192000 log 22.4125
24! 620448401733239439360000 = 2^22 3^10 5^4 7^3 11^2 13 17 19 23 div 242880 log 23.7927
25! 15511210043330985984000000 = 2^22 3^10 5^6 7^3 11^2 13 17 19 23 div 340032 log 25.1906
jagy@phobeusjunior:~$ ./Superior_Highly_Composite_read
2 = 2 div 2 log 0.30103
6 = 2 3 div 4 log 0.778151
12 = 2^2 3 div 6 log 1.07918
60 = 2^2 3 5 div 12 log 1.77815
120 = 2^3 3 5 div 16 log 2.07918
360 = 2^3 3^2 5 div 24 log 2.5563
2520 = 2^3 3^2 5 7 div 48 log 3.4014
5040 = 2^4 3^2 5 7 div 60 log 3.70243
55440 = 2^4 3^2 5 7 11 div 120 log 4.74382
720720 = 2^4 3^2 5 7 11 13 div 240 log 5.85777
1441440 = 2^5 3^2 5 7 11 13 div 288 log 6.1588
4324320 = 2^5 3^3 5 7 11 13 div 384 log 6.63592
21621600 = 2^5 3^3 5^2 7 11 13 div 576 log 7.33489
367567200 = 2^5 3^3 5^2 7 11 13 17 div 1152 log 8.56534
6983776800 = 2^5 3^3 5^2 7 11 13 17 19 div 2304 log 9.84409
13967553600 = 2^6 3^3 5^2 7 11 13 17 19 div 2688 log 10.1451
321253732800 = 2^6 3^3 5^2 7 11 13 17 19 23 div 5376 log 11.5068
2248776129600 = 2^6 3^3 5^2 7^2 11 13 17 19 23 div 8064 log 12.3519
65214507758400 = 2^6 3^3 5^2 7^2 11 13 17 19 23 29 div 16128 log 13.8143
195643523275200 = 2^6 3^4 5^2 7^2 11 13 17 19 23 29 div 20160 log 14.2915
6064949221531200 = 2^6 3^4 5^2 7^2 11 13 17 19 23 29 31 div 40320 log 15.7828
12129898443062400 = 2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 div 46080 log 16.0839
448806242393308800 = 2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 div 92160 log 17.6521
18401055938125660800 = 2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 div 184320 log 19.2648
791245405339403414400 = 2^7 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 43 div 368640 log 20.8983
Made one just a little higher but without the factorizations, just the number of divisors and the logarithm base ten.
jagy@phobeusjunior:~$ ./factorial
2! 2 div 2 log 0.30103
3! 6 div 4 log 0.778151
4! 24 div 8 log 1.38021
5! 120 div 16 log 2.07918
6! 720 div 30 log 2.85733
7! 5040 div 60 log 3.70243
8! 40320 div 96 log 4.60552
9! 362880 div 160 log 5.55976
10! 3628800 div 270 log 6.55976
11! 39916800 div 540 log 7.60116
12! 479001600 div 792 log 8.68034
13! 6227020800 div 1584 log 9.79428
14! 87178291200 div 2592 log 10.9404
15! 1307674368000 div 4032 log 12.1165
16! 20922789888000 div 5376 log 13.3206
17! 355687428096000 div 10752 log 14.5511
18! 6402373705728000 div 14688 log 15.8063
19! 121645100408832000 div 29376 log 17.0851
20! 2432902008176640000 div 41040 log 18.3861
21! 51090942171709440000 div 60800 log 19.7083
22! 1124000727777607680000 div 96000 log 21.0508
23! 25852016738884976640000 div 192000 log 22.4125
24! 620448401733239439360000 div 242880 log 23.7927
25! 15511210043330985984000000 div 340032 log 25.1906
26! 403291461126605635584000000 div 532224 log 26.6056
27! 10888869450418352160768000000 div 677376 log 28.037
28! 304888344611713860501504000000 div 917280 log 29.4841
29! 8841761993739701954543616000000 div 1834560 log 30.9465
30! 265252859812191058636308480000000 div 2332800 log 32.4237
31! 8222838654177922817725562880000000 div 4665600 log 33.915
32! 263130836933693530167218012160000000 div 5529600 log 35.4202
jagy@phobeusjunior:~$ ./Superior_Highly_Composite_read
2 div 2 log 0.30103
6 div 4 log 0.778151
12 div 6 log 1.07918
60 div 12 log 1.77815
120 div 16 log 2.07918
360 div 24 log 2.5563
2520 div 48 log 3.4014
5040 div 60 log 3.70243
55440 div 120 log 4.74382
720720 div 240 log 5.85777
1441440 div 288 log 6.1588
4324320 div 384 log 6.63592
21621600 div 576 log 7.33489
367567200 div 1152 log 8.56534
6983776800 div 2304 log 9.84409
13967553600 div 2688 log 10.1451
321253732800 div 5376 log 11.5068
2248776129600 div 8064 log 12.3519
65214507758400 div 16128 log 13.8143
195643523275200 div 20160 log 14.2915
6064949221531200 div 40320 log 15.7828
12129898443062400 div 46080 log 16.0839
448806242393308800 div 92160 log 17.6521
18401055938125660800 div 184320 log 19.2648
791245405339403414400 div 368640 log 20.8983
37188534050951960476800 div 737280 log 22.5704
185942670254759802384000 div 983040 log 23.2694
9854961523502269526352000 div 1966080 log 24.9937
581442729886633902054768000 div 3932160 log 26.7645
1162885459773267804109536000 div 4423680 log 27.0655
12791740057505945845204896000 div 6635520 log 28.1069