Leaving aside the trivial cases $n=0$ and $n=1$, assume $n \ge 2$. The conclusion follows from the fact that the polynomial under consideration,

$$f = \sum_{k=0}^{n} \binom{n}{k}^2 (X+1)^{2k} (X-1)^{2(n-k)},$$
is expressible as a sum of $n+1$ polynomials of the form $(b_k X^2 + c_k)^n$, where the $b_k$ and the $c_k$ are all non-negative real numbers and at least $n-1$ products $b_k c_k$ are positive; this latter then implies that the even powers of $X$ all occur in the expansion with a positive coefficient. In particular, the fact that the odd rank coefficients in the standard expansion all vanish comes for free.

Let $S$ be the set of $(n+1)$-st roots of unity and recall that, if $k$ and $\ell$ are integers in the range $0$ through $n$, then the sum $\sum_{\omega \in S} \omega^k \bar{\omega}^\ell$ vanishes, unless $k=\ell$ in which case it is equal to $n+1$.

$$\begin{align*} f &= \frac{1}{n+1} \sum_{k=0}^{n} \sum_{\ell=0}^{n} \binom{n}{k} \binom{n}{\ell} (X+1)^{k+\ell} (X-1)^{2n-k-\ell} \sum_{\omega \in S} \omega^k \bar{\omega}^{\ell} \\ &= \frac{1}{n+1} \sum_{\omega \in S} \left( \sum_{k=0}^{n} \binom{n}{k} \omega^k (X+1)^k (X-1)^{n-k} \right) \\ &\qquad \cdot \left( \sum_{\ell=0}^{n} \binom{n}{\ell} \bar{\omega}^{\ell} (X+1)^{\ell} (X-1)^{n-\ell} \right) \\ &= \frac{1}{n+1} \sum_{\omega \in S} (\omega(X+1) + X-1)^n (\bar{\omega}(X+1) + X-1)^n \\ &= \frac{1}{n+1} \sum_{\omega \in S} \left( (2+\omega+\bar{\omega})X^2 + 2 - \omega - \bar{\omega} \right)^n. \end{align*}$$

Finally, since each $\omega + \bar{\omega}$ is a real number whose absolute value does not exceed $2$ and $\omega + \bar{\omega} < 2$ for at least $n-1$ roots $\omega \neq \pm 1$, each summand above is a polynomial of the desired form. This completes the proof.