As for your polynomial $f=\prod_{i=1}^5 (x_i+x_{i+1})^2$, you may do the following trick. At first, you replace your polynomial to $g=\prod_{i=1}^5 (x_i+x_{i+1}-4)(x_i+x_{i+1}-3)$. Antisymmetrizations of $f$ and $g$ are the same, since their difference has degree less than 10. Next, we use points 0,1,2,3,4 as $x$'s. I think, there is unique cyclic permutation 10234 for which $g$ is non-zero. Hence all non-zero summands in the sum $\sum {\rm\,sign}\,(\pi) g(x_{\pi_1},x_{\pi_2},\dots,x_{\pi_5})$ are equal, so this sum is non-zero.

As for your polynomial $f=\prod_{i=1}^5 (x_i+x_{i+1})^2$, you may do the following trick. At first, you replace your polynomial to $g=\prod_{i=1}^5 (x_i+x_{i+1}-4)(x_i+x_{i+1}-3)$. Antisymmetrizations of $f$ and $g$ are the same, since their difference has degree less than 10. Next, we use points 0,1,2,3,4 as $x$'s. I think, there is unique (up to dihedral group symmetry) permutation 10234 for which $g$ is non-zero. Hence all non-zero summands in the sum $\sum {\rm\,sign}\,(\pi) g(x_{\pi_1},x_{\pi_2},\dots,x_{\pi_5})$ are equal, so this sum is non-zero.