I am looking for (approximate) closed forms for the following expression: $$ {n \choose n/2} p^n = \frac{n!}{(n/2)!(n/2)!} p^n, $$ for a very small constant $p$ (say $p = 1e-5$).
What if $p$ is a function of $n$? (say $ p = 1/n$ or $p = \frac{\log n}{n}$)
On
It could interesting for you to look at robjohn's answer to this question $$\frac{4^m}{\sqrt{\pi(m+\frac13)}}\le\binom{2m}{m}\le\frac{4^m}{\sqrt{\pi(m+\frac14)}}$$ which makes $$\frac{2^n \sqrt 6}{\sqrt{\pi(3n+2)}}\le\binom{n}{\frac n 2}\le\frac{2^{n+1}}{\sqrt{\pi(2n+1)}}$$ which are extremely good bounds for your particular central binomial coefficient as shown in the table below $$\left( \begin{array}{cccc} n & \text{left bound} & \text{exact}& \text{right bound}\\ 5 & 10.7257 & 10.865 & 10.887 \\ 10 & 250.164 & 252. & 252.142 \\ 15 & 6605.44 & 6639.13 & 6640.86 \\ 20 & 184037. & 184756. & 184783. \\ 25 & 5.28452\times 10^6 & 5.30125\times 10^6 & 5.30176\times 10^6 \\ 30 & 1.54706\times 10^8 & 1.55118\times 10^8 & 1.55128\times 10^8 \\ 35 & 4.59049\times 10^9 & 4.60102\times 10^9 & 4.60125\times 10^9 \\ 40 & 1.37569\times 10^{11} & 1.37847\times 10^{11} & 1.37852\times 10^{11} \\ 45 & 4.15423\times 10^{12} & 4.16170\times 10^{12} & 4.16183\times 10^{12} \\ 50 & 1.26206\times 10^{14} & 1.26411\times 10^{14} & 1.26414\times 10^{14} \end{array} \right)$$ I am sure that you could "play" a lot with these.
You can approximate the central binomial coefficient $\binom{2n}{n} \approx \frac{4^n}{\sqrt{\pi n}} $ using Stirling's formula (the approximation is in the sense that $\lim_{n \to \infty} \frac{ \frac{4^n}{\sqrt{\pi n}} } {\binom{2n}{n} } =1$). Wikipedia link gives you some more information on the error in the approximation if you need it.
You can write $p^n = 4^{n \log_4 p}$ and combine it with the approximation given above to get an approximation when $p$ is a function of $n$.