To give a solution from scratch, start with the identity
$$\sum_{s=-\infty}^\infty\frac1{(x+s)^2}=\frac{\pi^2}{\sin^2\pi x}.$$
We want to prove that this extends to an identity
$$\sum_{s=-\infty}^\infty\frac1{(x+s)^n}=\frac{\pi^n P_n(\cos\pi x)}{\sin^n\pi x}$$
for each integer $n\ge2$, where the $P_n$ are polynomials satisfying some
nice recursion. If this holds for some $n$ then differentation yields
$$\sum_{s=-\infty}^\infty\frac1{(x+s)^{n+1}}
=\frac{\pi^{n+1}\cos\pi x\\, P_n(\cos\pi x)}{\sin^{n+1}\pi x}
+\frac{\pi^{n+1}\sin^2\pi x\\,P_n'(\cos\pi x)}{n\sin^{n+1}\pi x}$$
giving
$$P_{n+1}(t)=tP_n(t)+\frac{(1-t^2)}{n}P_n'(t).$$
So, in your notation, $P_n=Q_{n-2}$.

However
$$\sum_{s=-\infty}^\infty\mathrm{sinc}(\pi(x+s))^n
=\sum_{s=-\infty}^\infty\frac{(-1)^{sn}\sin^n\pi x}{\pi^n(x+s)^n}
=\frac{\sin^n\pi x}{\pi^n}\sum_{x=-\infty}^\infty\frac{(-1)^{ns}}{(x+s)^n}$$
which is the above sum when $n$ is even but not when $n$ is odd. Unless
I have made some sign error (quite likely!) then your assertion holds for
odd $p$ but needs to be modified for even $p$. Note that
$$\sum_{s=-\infty}^\infty\frac{(-1)^n}{(x+s)^n}=
2^{1-n}\sum_{s=-\infty}^\infty\frac1{(x/2+s)^n}
-\sum_{s=-\infty}^\infty\frac1{(x+s)^n}$$
so this is feasible.