Let $n\geq 2$ be an integer and let $\Sigma$ be the collection of all $2$-subsets (a 2-set is a set that contains $2$ elements) of $[n]=\{1,2,\dots,n\}$, thus $\Sigma$ contains $\binom{n}{2}$ elements. Let $J_n$ be the $\Sigma\times\Sigma$-matrix such that $J_n(S,S')=x_{|S\Delta S'|}$ for $S,S'\in\Sigma$, calculate $\det(J_n)$.
For example, if $n=3$, then $\Sigma=\{\{1,2\},\{1,3\},\{2,3\}\}$ and $$J_3=\begin{pmatrix} x_0 & x_2 & x_2\\ x_2 & x_0 & x_2\\ x_2 & x_2 & x_0 \end{pmatrix},$$ thus we have $\det(J_3)=(x_0-x_2)^2(x_0+2x_2)$.
If $n=4$, then $\Sigma=\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\}$ and $$J_4=\begin{pmatrix} x_0 & x_2 & x_2 & x_2 & x_2 & x_4\\ x_2 & x_0 & x_2 & x_2 & x_4 & x_2\\ x_2 & x_2 & x_0 & x_4 & x_2 & x_2\\ x_2 & x_2 & x_4 & x_0 & x_2 & x_2\\ x_2 & x_4 & x_2 & x_2 & x_0 & x_2\\ x_4 & x_2 & x_2 & x_2 & x_2 & x_0\\ \end{pmatrix},$$ thus we have $$\det(J_4)=(x_0-x_4)^3(x_0-2x_2+x_4)^2(x_0+4x_2+x_4).$$ Similarily, we have $$\det(J_5)=(x_0-2x_2+x_4)^5(-x_0-x_2+2x_4)^4(x_0+6x_2+3x_4).$$ These results are a little messy, how can we find a general formula for all $n\geq 2$? It seems that we can use recursion.