Do most numbers have exactly $3$ prime factors?

Yes, the line for numbers with $3$ prime factors will be overtaken by another line. As shown & explained in Prime Factors: Plotting the Prime Factor Frequencies, even up to $10$ million, the most frequent count is $3$, with the mean being close to it. However, it later says

For $n = 10^9$ the mean is close to $3$, and for $n = 10^{24}$ the mean is close to $4$.

The most common # of prime factors increases, but only very slowly, and with the mean having "no upper limit".

OEIS A$001221$'s closely related (i.e., where multiplicities are not counted) Number of distinct primes dividing n (also called omega(n)) says

The average order of $a(n): \sum_{k=1}^n a(k) \sim \sum_{k=1}^n \log \log k.$ - Daniel Forgues, Aug 13-16 2015

Since this involves the log of a log, it helps explain why the average order increases only very slowly.

In addition, the Hardy–Ramanujan theorem says

... the normal order of the number $\omega(n)$ of distinct prime factors of a number $n$ is $\log(\log(n))$.

Roughly speaking, this means that most numbers have about this number of distinct prime factors.

Also, regarding the statistical distribution, you have the Erdős–Kac theorem which states

... if $ω(n)$ is the number of distinct prime factors of $n$ (sequence A001221 in the OEIS, then, loosely speaking, the probability distribution of

$$\frac {\omega (n)-\log \log n}{\sqrt {\log \log n}}$$

is the standard normal distribution.

To see graphs related to this distribution, the first linked page of Prime Factors: Plotting the Prime Factor Frequencies has one which shows the values up to $10$ million.

Just another plot to about $250\times10^9$, showing the relative amount of numbers below with x factors (with multiplicity)

Somewhere between $151,100,000,000$ and $151,200,000,000$ 4 overtakes r3.