With slightly different notations, we look for a coefficient of $\prod a_i^2b_i^2c_i^2$ in the polynomial $\prod_{i=1}^{2n} (a_{i+1}-a_i)(b_{i+1}-b_i)(c_{i+1}-c_i)\cdot \prod_{i=1}^{2n} (a_i-b_i)(b_i-c_i)(c_i-a_i)$, where indices are taken modulo $n$. We may apply [this][1] formula for all sets $A_i=\{0,1,2\}$. How does a non-zero value occur? For each $i$, $[a_i,b_i,c_i]$ is a permutation $\pi_i$ of $0,1,2$, and $\pi_{i+1}$ is obtained from $\pi_i$ be a cyclic shift (there are two possible shifts). If we introduce $\varepsilon_i=\pm 1$ dependently on what shift is used, then $\sum \varepsilon_i$ must be divisible by and the value of the polynomial is proportional to $\prod \varepsilon_i$. So the coefficient equals $$6\sum_{3|\varepsilon_1+\dots+\varepsilon_{2n}}\varepsilon_1\cdot \ldots \cdot\varepsilon_{2n}.$$
Thу sum equals $\frac13((1-1)^{2n}+w^{-n}(1-w)^{2n}+w^{-2n}(1-w^2)^{2n})=\frac23 (-3)^n$, where $w=e^{2\pi i/3}$, thus the answer is $4(-3)^n$.


  [1]: https://mathoverflow.net/a/214930/4312