Since $2^{-n}\binom{n}{k}$ has mean $\frac n2$ and variance $\frac n4$
$$
2^{-n}\binom{n}{k}\sim\frac1{\sqrt{\pi n/2}}e^{-2(k-n/2)^2/n}
$$
Therefore,
$$
\begin{align}
\frac1{n^p}\sum_{k=0}^n(n-2k)^{2p}2^{-n}\binom{n}{k}
&\sim\frac1{n^p}\frac1{\sqrt{\pi n/2}}\int_{-\infty}^\infty(2x)^{2p}e^{-2x^2/n}\,\mathrm{d}x\\
&=\frac2{\sqrt{2\pi}}\int_0^\infty x^{2p}e^{-x^2/2}\,\mathrm{d}x\\
&=\frac2{\sqrt{2\pi}}\int_0^\infty x^{p-1/2}e^{-x/2}\cdot\tfrac12\mathrm{d}x\\
&=\frac{2^p}{\sqrt{\pi}}\int_0^\infty x^{p-1/2}e^{-x}\,\mathrm{d}x\\