This question is similar to one concerning nested logarithms and also nested radicals involving factorials. If I am correct a quick calculation shows:
$$\ln (1! \ln (2! \ln (3! \ln (4! \ln (5! \ln (6! \ldots ))))))\approx 0.654569$$
Does the sequence implied by this infinite nested logarithm converge to a finite value and if so to what value exactly?
Consider $$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\dots}}}}$$ We have that $n!^{2^{-x}} \to 1$, which is sufficient to conclude the above converges. In other words, we have that $$f(1!f(2!f(3!f(\dots))))$$ Where $f(x)=\sqrt{x}$, converges. Noting that $g(x)=\ln(x)$ is positive but less than $f(x)$ for all $x>0$, it follows that $$g(1!g(2!g(3!g(\dots))))$$ also converges. The notation is a little icky, but you could also define it recursively. I checked the number computed by user Peter in the comments in OEIS and through an inverse symbolic calculator, neither of which turned up any results. If you wanted a closed form for what it converges to, you are not likely to find one, though its convergence is rapid and decimal approximations become very accurate very quickly.
Edit: slight mistake. That $n!^{2^{-x}}$ converges is sufficient to conclude $$\sqrt{1 + \sqrt{2! + \sqrt{3! + \dots}}}$$ converges, but it is not sufficient to conclude $$\sqrt{1!\sqrt{2!\sqrt{\dots}}}$$ converges. Determining whether the above converges is synonymous to showing $$\prod_{n=1}^\infty n!^{2^{-x}}$$ Converges. I will not show this, but you can take my word that it does converge. So, because the above converges, we have that $\sqrt{1\sqrt{2\dots}}$ converges. Because $\sqrt{x} > \ln(x)$, your infinitely nested logarithm must be bounded by $\sqrt{1!\sqrt{2!\dots}}$, and therefore must also converge.
