Solution:
Answer: $764$
Let $a_{n}$ be the sum of the coefficients of the square-terms in the product $\prod_{1 \leq i<j \leq n}(1+ x_{i} x_{j})$. Square-free terms in this product come in two types: either they include $x_{n}$, or they do not. The sum of the coefficients of the terms that include $x_{n}$ is $(n-1) a_{n-2}$, since we can choose any of the $n-1$ other variables to be paired with $x_{n}$, and then choose any square-free term from the product taken over the other $n-2$ variables. The sum of the coefficients of the terms that do not include $x_{n}$ are $a_{n-1}$, because they all come from the product over the other $n-1$ variables. Therefore, $a_{n}=a_{n-1}+(n-1) a_{n-2}$.

We use this recursion to find $a_{8}$. As base cases, $a_{0}$ and $a_{1}$ are both $1$. Then $a_{2}=2$, $a_{3}=4$, $a_{4}=10$, $a_{5}=26$, $a_{6}=76$, $a_{7}=232$, and finally, $a_{8}=764$.