Notice that this sum equals 
$$\frac1{2n} \sum_{k=1}^n (-1)^k k^{p} \binom{2n}{n+k}=\frac1{4n} \sum_{k=0}^{2n} (-1)^k (k-n)^{p} \binom{2n}{k}=\frac1{4n}\left.\Delta^{2n}x^p\right|_{x=-n},$$
which is zero for even $p<2n$.