The easiest way to answer this question is with the steepest descent method, which is a standard techinque for calculating such asymptotic expansions. Here is the case $p(x)=x^2$. Write the sum as
$$ S = \sum_{0\leq k\leq n} \binom{p(k)}{n-k} = \sum_{0\leq k\leq n} s(k). $$

For the binomial coefficients one useful asymptotic formula is
$$ \binom{n}{n y} = q(n,y) = (2\pi n y(1-y))^{-1/2}\exp(n H(y)), $$
where $H(y) = -y\log y-(1-y)\log(1-y)$ is the binary entropy function.

We can approximate the sum by the equivalent integral, and then expand the function $s(k)$ in Taylor series around its maximum $s(k_*)=s_*$:
$$ \begin{aligned}
S &\sim \int_0^n s(k)\,dk = n \int_0^1 s(n z)\,dz
\\&\sim n \int_{-\infty}^\infty s_* e^{\frac12\partial_z^2(\log s)(z-z_*)^2}\,dz
\\&= s_* n \sqrt{2\pi/(-\partial_z^2(\log s))}.
\end{aligned} $$

The find the point $z_*$ and the corresponding term $-\partial_z^2(\log s)$, it is sufficient to approximate $s(n z)$ by
$$ s(n z) \sim q(p(n z), n(1-z)/p(n z)) = q\left(n^2z^2,\frac{1-z}{z^2n}\right). $$
To find the leading asymptotic term in $z$, it is easiest to use $\log q$ instead of $q$:
$$\begin{aligned}
\log q &= n^2 z^2 H\left(\frac{1-z}{n z^2}\right) \sim n(1-z)\log\frac{e n z^2}{1-z}, \\
\partial_z \log q &\sim n\left(-2+\frac2z - \log n + \log\frac{1-z}{z^2} \right)
\\
\partial_z^2 \log q &\sim -\frac{n(2-z^2)}{z^2(1-z)}.
\end{aligned}$$
Setting $\partial_z \log q=0$, and substituting $z=1/w$ gives the equation
$$ w + \log w + \frac12\log(1-1/w) = \frac12 \log e^2n, $$
which can be written in this form:
$$ w e^w = \frac{e \sqrt{n}}{\sqrt{1-1/w}}. $$
To leading order, $w$ is given by the Lambert W function:
$$ w = W(e \sqrt{n}), $$
and the next approximation is:
$$ w = W(e \sqrt{n})/\sqrt{1-1/W(e \sqrt{n})}. $$
From $z=1/w$ it is also possible to calculate the asymptotic form for $s_*$.

So the result is that
$$ \frac{S}{s_*} \sim n\sqrt{2\pi}\left(\frac{n W(e\sqrt{n})(-1+2W(e \sqrt n)^2)}{W(e \sqrt{n})-1} \right)^{-1/2}. $$
At $n=100$, the error is $6.6\%$, and the next order for $w$ gives an error of $0.69\%$. Asymptotically, $W(e \sqrt n) \sim \log(e\sqrt n)$, but the convergence is slow, at $n=100$ the error is $-16.6\%$.