With slightly different notations (and with the opposite sign), we look for the coefficient of $\prod a_i^2b_i^2c_i^2$ in the polynomial $F:=\prod_{i=1}^{2n} (a_{i+1}-a_i)(b_{i+1}-b_i)(c_{i+1}-c_i)\cdot \prod_{i=1}^{2n} (a_i-b_i)(b_i-c_i)(c_i-a_i)$, 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 of $F$ 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 $\pi_i$ by a cyclic shift (there are two possible shifts). If we introduce $\varepsilon_i=\pm 1$ dependently on which shift is used, then $\sum \varepsilon_i$ must be divisible by 3 and the value of the polynomial is $4^{2n}\prod \varepsilon_i$. The denominator in the formula is always $4^{2n}$. So the coefficient equals $$6\sum_{3|\varepsilon_1+\dots+\varepsilon_{2n}}\varepsilon_1\cdot \ldots \cdot\varepsilon_{2n}=6\sum_{3 |m-n} (-1)^m\binom{2n} {m}, $$
6 corresponds to the choice of $\pi_1$, $m$ denotes the number of negative $\varepsilon$'s.
The 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 $4(-3)^n$.

Here we use the fact (somebody here on MO told how is it called but I forgot) that for a Laurent polynomial $f(z)=\sum a_j z^j$ and a positive integer $n$ we have $$\sum_{k\equiv \alpha\pmod n} a_k=\frac1n\sum_{\omega:\omega^n=1} \omega^{-\alpha}f(\omega).$$


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