I've found this sum:
$$S(n) = \sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$$
The inner sum is elliptical iirc, but perhaps the double sum has a nice expression. We can change the order of summation and the expression becomes a tiny bit better to manipulate, but I went straight to try some induction on it. So here is the sequence $S(n)$ as $n$ goes from $1$ to $10$:
$$-1,5,-61,614,-5762, 52093, -460099, 3999356,-34363396, 292676603,...$$
I tried some polynomial interpolations but they didn't stabilize on the values I tried (something like $n=6$). Then I did suppose $S(n) = (-1)^n a^n p(n)$ for $p(n)$ a polynomial and $a=2$ or $a=4$, it also didn't stabilize. I have good reasons to believe
$$S(n) = a \frac{(-1)^n}{n+d} { 4n+b \choose n+c}$$
for some constants $a,b,c$ and $d$ but I'm not sure how to find those constants. Any ideas on how to solve this problem?
EDIT: I made a Python program that verifies those constants and I got nothing for each on the interval $[0,19]$, perhaps $S(n) = \mathcal{o}((\frac{256}{27})^n\frac1{n \sqrt n})$, indeed $\frac1n{4n \choose n}$ begin dominating a bit after $n=40$
EDIT2: changing order of summation: $$n \le k \le i \le 2n-1 \implies S(n) = \sum_{i=n}^{2n-1} \sum_{k=n}^i {2n \choose i+1} {i \choose n}k(-1)^k = \\ = \sum_{i=n}^{2n-1} {2n \choose i+1} {i \choose n} \sum_{k=n}^i (-1)^k k = \frac14 \sum_{i=n}^{2n-1} {2n \choose i+1} {i \choose n}[(-1)^i(2i+1) + (-1)^n(2n-1)] $$