Considering $$A_n={2n \choose n - n \epsilon}=\frac{(2n)!}{(n-n\epsilon)! \,(n+n\epsilon)!}$$ replace the factorials by the corresponding Gamma functions. Taking logarithms and expanding as Taylor series around $\epsilon=0$, one could obtain $$\log(A_n)=(\log ((2 n)!)-2 \log (n!))-n^2 \epsilon ^2 \psi
^{(1)}(n+1)+O\left(\epsilon ^4\right)$$ where appears the first derivative of the digamma function.
Using $\epsilon=\frac 1 {\log(n)}$ and plotting, it seems to be a quite good approximation as show in the table below
$$\left(
\begin{array}{ccc}
n & \log(A_n) & \text{approx} \\
5 & 3.67380 & 3.77940 \\
10 & 10.2770 & 10.3318 \\
15 & 16.8366 & 16.8810 \\
20 & 23.4355 & 23.4756 \\
25 & 30.0673 & 30.1053 \\
30 & 36.7243 & 36.7611 \\
35 & 43.4009 & 43.4370 \\
40 & 50.0931 & 50.1288 \\
45 & 56.7981 & 56.8336 \\
50 & 63.5137 & 63.5492
\end{array}
\right)$$
For sure, we could make better increasing the order of the expansion to give
$$\log(A_n)=(\log ((2 n)!)-2 \log (n!))-n^2 \epsilon ^2 \psi
^{(1)}(n+1)-\frac{1}{12} n^4 \epsilon ^4 \psi ^{(3)}(n+1)+O\left(\epsilon ^6\right)$$ The next table reproduces some results for this secod approximation.
$$\left(
\begin{array}{ccc}
n & \log(A_n) & \text{approx} \\
5 & 3.67380 & 3.68759 \\
10 & 10.2770 & 10.2809 \\
15 & 16.8366 & 16.8389 \\
20 & 23.4355 & 23.4372 \\
25 & 30.0673 & 30.0688 \\
30 & 36.7243 & 36.7256 \\
35 & 43.4009 & 43.4020 \\
40 & 50.0931 & 50.0941 \\
45 & 56.7981 & 56.7990 \\
50 & 63.5137 & 63.5146
\end{array}
\right)$$