Too long for a comment.
Making the problem more general and using the error bound given in @Gary's comment, you are looking for $n$ such that
$$\frac{1}{{2^n (n + 1)!}}\left( {\frac{\pi }{4}} \right)^{n + 1} \leq 10^{ - k}$$ that is to say
$$(n+1)! \geq 2 \left(\frac{\pi}{8}\right)^{n+1} 10^k$$
If you look at this question of mine, you will see a magnificent approximation by @robjohn.
Applied to to this problem
$$\color{blue}{n \sim \frac \pi 8 e^{1+W(t)}-\frac 3 2} \qquad \text{where} \qquad \color{blue}{t=\frac{4}{e \pi }\log \left(\frac{8 \times 10^{2 k} }{\pi ^2}\right)}$$ where $W(t)$ is Lambert function. As usual, you will use $\lceil n \rceil$.
Using it for $k=32$, this will give, as a real, $n=22.3097$ while the exact solution is $n=22.3101$. Checking
$$\frac{1}{{2^{22}\, 23!}}\left( {\frac{\pi }{4}} \right)^{23}=3.56\times 10^{-32} > 10^{-32}$$
$$\frac{1}{{2^{23}\, 24!}}\left( {\frac{\pi }{4}} \right)^{24}=5.83\times 10^{-34} < 10^{-32}$$