Computational results for the sequence $n!+{p_n}!+1$ are, well, very very unusual

Peter and I were discussing in a chat room and I thought that it would be nice to test the sequence $$n!+{p_n}!+1$$ for primality.

Then I wrote Peter that I expect much of primes in this sequence, well, at least the expression is formed in such a way that there could be much of them.

But, to my and Peter´s surprise, he computed and tested this sequence for all $1\leq n \leq 100$ and found that for only $n=3$ in that range the expression is prime.

He is pushing his computations even further, even at this moment, and, for me almost unbelievable is that he passed $n=800$ and found that only for $n=3$ the expression is prime and we are curious why there is (so far) only one prime although small factors are impossible for large values of $n$.

So, we agreed that a question about this unusuality should be asked, and I ask three of them, very much interrelated and connected and answerable.

Do we have any explanation of why this sequence has, for the range computed, so a small number of primes? Is it naive to expect that this sequence has an infinite number of primes? What "should" be expected to happen in some very large ranges?

Update: No further primes up to $n=1000$.

Maybe not enough for a full answer but nonetheless some remarks. If the number $n!+1$ has some small prime factor (i.e. less than $p_n$) that immediately implies that $n!+p_n!+1$ is divisible by that prime as well and therefore not a prime number. Unfortunately, it is even not known whether or not there are infinitely many primes among $n!+1$ which btw is OEIS A002981. It could be likely that $n$ for which it is prime would be contained in this sequence - $n=3$ is in this sequence too. I've checked myself $n=872$ and $n=1477$ but unfortunately both of them are composite. Also if for some $n$ from this sequence the number is composite, its prime factors tend to be large, for example for $n=11$ we have: $$11!+31!+1=192119825921 \cdot 42800573104616324673281$$ Which makes it harder to prove anything by using congruences. I am now in the middle of checking $n=6380$. However: $$6380! + 63647! + 1 \approx 2 \cdot 10^{278107}$$ Is a huge number and it will take hours to check its primality using my old computer. Maybe someone with better access to the computing power can handle it faster?
Also I would like to point out that the heuristic from the comments should be somewhat improved (but it still provides much insight into the problem) - we shouldn't consider only the probability of being prime (which we can say is roughly $\log n$) but rather take the probability of $m=n!+p_n+1$ being prime and conditioned on the fact that it is not divisible by any prime $\le n$ - it will make the probability higher and can make our sum divergent (however I am not too certain about it, it's just an idea - I don't even know how one would go about computing such conditioned probability in a meaningful way).

EDIT

I've recently checked that: $$6380! + 63647! + 1 \approx 2 \cdot 10^{278107}$$ Is composite. Also, by Peter's comment if we denote by $A$ the event that: $$m=n!+p_n!+1$$ Is prime and by $B$ the event that it has no prime less than $n$ we have $A \subset B$ and that asymptotically: $$P(A)=\frac{1}{\log m}=\frac{1}{n \log n^2}$$ $$P(B)=\frac{e^{-\gamma}}{\log n}$$ And therefore: $$P(A \mid B) = \frac{P(A)}{P(B)} = \frac{e^{\gamma}}{n \log n}$$ Which is a divergent series therefore indicating that there can be infinitely many such numbers. However, the sum diverges very slowly: $$\sum_{n=10}^{1000} \frac{e^{-\gamma}}{n \log n}\approx 2$$ So we shouldn't expect such primes too often. Actually, we have that since: $$\int \frac{1}{x \log x}=\log \log x + C$$ We can write: $$\sum_{n=2}^{N} \frac{e^{-\gamma}}{n \log n} = O(\log \log N)$$ So we shouldn't expect many of them even for big values of $N$ and it's hard to check them because the numbers grow so rapidly. I am now in the middle of checking the next number from the A002981 i.e. $n=26951$. However, since: $$26951! + 312029! + 1 \approx 6.8 \cdot 10^{1578838}$$ It will take "a bit" longer for my computer to check its primality.

EDIT2

After three days of computation I still haven't managed to check for the primality so I've decided to finally give up.