You may also prove ${2n\choose n}\sim \frac{4^n}{\sqrt{\pi n}}$ using the combinatorial meaning of ${2n\choose n}$ and the area-of-a-circle definition of $\pi$. That is probably the most elementary we can hope for.

First of all, if $a_n:=\sqrt{n+1/4}{2n\choose n}4^{-n}$ and $b_n:=\sqrt{n+1/2}{2n\choose n}4^{-n}$, we may check that $a_n\leqslant b_n$, $a_n$ increases and $b_n$ decreases, thus $a_n$, $b_n$ have a common positive finite limit which we denote $\alpha$, and $a_n\leqslant \alpha\leqslant b_n$. Since $b_n/a_n=1+O(1/n)$, we get $a_n=\alpha+O(1/n)$, and also $c_n=\alpha+O(1/n)$ where $c_n:=\sqrt{n}{2n\choose n}4^{-n}$.

We should prove that $\alpha=1/\sqrt{\pi}$.

Use the combinatorial identity $$4^{-n}\sum_{k=0}^n k{2k\choose k}(n-k){2(n-k)\choose n-k}=\frac{n(n-1)}8.\quad (\heartsuit)$$
LHS of $(\heartsuit)$ equals
$$
\sum_{k=0}^n \sqrt{k(n-k)}c_kc_{n-k}=\sum_{k=0}^n \sqrt{k(n-k)}\left(\alpha^2+O\left(\frac1k+\frac1{n-k}\right)\right)\\=\alpha^2\sum_{k=0}^n\sqrt{k(n-k)}+O\left(n\sum_{k=0}^n\frac1{\sqrt{k(n-k)}}\right)=\alpha^2\cdot \frac{\pi n^2}8+o(n^2),
$$
since the union of rectangles $[k,k+1]\times [0,\sqrt{k(n-k)}]$ approximates the upper semi-circle with the horizontal diameter $[0,n]$.

Thus $\alpha^2\pi/8=1/8$ and $\alpha=1/\sqrt{\pi}$.

You could use instead of $(\heartsuit)$ the more famous identity
$$
4^{-n}\sum_{k=0}^n {2k\choose k}{2(n-k)\choose n-k}=1
$$
this gives $\alpha^2\int_0^1\frac{dx}{\sqrt{x(1-x)}}=1$ (the integral equals $\pi$ of course, but this is less "elementary" then the area of a circle.)

This argument is similar to J. Wästlund's
(An elementary proof of Wallis' product formula for $\pi$, 2007).