Behavior of $\sum_{k=0}^n (-1)^k {2n \choose k} (1-\frac{k}{n})^2 \ln(1-\frac{k}{n})$

As J. D'Aurizio notes, the proposer's question is answered if the asymptotics of his eq (1) can be determined. I'll outline a proof of the following: $$ \sum_{k=1}^n (-1)^k \binom{2n}{n+k} k^2 \log{k} \sim \binom{2n}{n} \frac{\pi}{4}\Big( 7\frac{\zeta(3)}{\pi^3} + \frac{93}{n} \frac{\zeta(5)}{\pi^5} + ...\Big).$$

The $\zeta(n)$ are Riemann's zeta. The OP conjectured an asymptotic form, which is correct up to an alternating factor, and asked what the value of the leading constant is. By using the asymptotics for the central binomial,

$$ \sum_{k=0}^n (-1)^k \binom{2n}{k} (1-k/n)^2 \log{(1-k/n)} \sim \frac{4^n}{n^{5/2}}\, (-1)^n \, \frac{7\zeta(3)}{4\pi^2\sqrt{\pi}}.$$

My top equation is not so difficult to derive if one should have the identity available,

$$ (A)\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = \binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}. $$ Differentiate with respect to $s$ and set it afterwards to 2. Use the asymptotic expansion for the ratio of gamma functions to get $$ \sum_{k=1}^n (-1)^{k} \binom{2n}{n+k} k^2 \log{k} = \binom{2n}{n} \frac{\pi}{2} \int_0^\infty \frac{dx \, \,x^2}{\sinh{\pi x}} \big(1+\frac{x^2}{n} + ...\big). $$ Integrate term-by-term and the integrals are known by Mathematica in terms of $\zeta(n)$.

I don't know of a reference that has (A). My proof is long and I'll skip it unless there is suitable interest. I was really surprised at the final result because I was expecting a $\log(n)$ to be in the mix.