For $N\rightarrow \infty$ I would like an approximation for $$ {{n}\choose{n/2-p}} - {{n}\choose{n/2-p-1}} $$ in particular around the point $p\approx n/4$.
It looks like the difference between two gaussians $$ \frac{2^{n+\frac{1}{2}} e^{-\frac{4 p (p+1)+2}{n}} \left(e^{\frac{2 (p+1)^2}{n}}-e^{\frac{2 p^2}{n}}\right)}{\sqrt{\pi } \sqrt{n}} $$ is not working really well. In particular, the accuracy does not seems to increase with $N$.
$$\binom{n}{k} - \binom{n}{k-1} = \frac{n!}{k!(n-k)!} - \frac{n!}{(k-1)!(n-k+1)!} \\ = \left(\frac{1}{k} - \frac{1}{(n-k+1)}\right)\frac{n!}{(k-1)!(n-k)!} \\ = \frac{n-2k+1}{k(n-k+1)} \frac{n!}{(k-1)!(n-k)!}$$
If you want to continue to try Gaussian approximation, you can rewrite this as $$\frac{n(n-2k+1)}{k(n-k+1)} \binom{n-1}{k-1}$$
Alternatively you can use Stirling's approximation to as many terms as you need.