Let $n$ and $p = 1/2$ be the parameters of a binomial distribution. We know that its probability mass function peaks at its expected value $n/2$ (suppose $n$ is even). Further, the probability at the peak can be expressed as
$$ {n \choose n/2} \frac{1}{2^n} = \frac{\prod_{a = n/2 + 1}^n a}{(\frac{n}{2})!} \cdot \frac{1}{2^n} $$
The question: it the resulting probability exponential in $n$ or polynomial in $n$?
There isn't a way to further simplify the exact probability.
In terms of asymptotics, you can use Stirling's approximation to get $\binom{n}{n/2} \sim \frac{2^n}{\sqrt{n\pi/2}}$ (e.g., see Wikipedia) so the probability behaves like $\sqrt{\frac{2}{\pi n}}$.
Connection to normal distribution: by a normal approximation, the $\text{Binom}(n, 1/2)$ distribution can be approximated by $N(n/2, n/4)$, whose density has a peak value of $\frac{1}{\sqrt{2\pi (n/4)}} = \sqrt{\frac{2}{\pi n}}.$