Almost certainly not. But note that if it were true, proving it would be extremely tough. We can't even prove now that there are infinitely many primes of the form $n!+1$ or that there are infinitely many primes of the form $n!-1$, and we can't prove that there are infinitely many twin primes, that is prime pairs which are two away from each other.
However, while we suspect that there are infinitely many in each of those three categories, we should strongly doubt that there are infinitely many $n$ where both $n!-1$ and $n!+1$ are prime.
Heuristically, the Prime Number Theorem tells that the "chance" a number x is prime should be roughly $\frac{1}{\log x}$. By Stirling's formula, $\frac{1}{\log n!}$ is very close to $\frac{1}{n \log n}$. So the expected number of such prime pairs should be $$\sum_{n=1}^{\infty} \frac{1}{\log (n!+1)} \frac{1}{\log (n!-1)} \approx \sum \frac{1}{(n \log n)^2},$$ which is a convergent series, so we should expect only finitely many such. I'd in fact go out on a limb and strongly suspect that $n=3$ is the largest $n$ where both $n!+1$ and $n!-1$ are prime. Using the relevant OEIS lists here and here one can see at a glance that there are no examples after $n=3$ below $n=150209$.