I. Let $n=2$ and $x = \frac1\phi$ with golden ratio $\phi$. Then it satisfies, $$x(1 - x^{12})^2(1 - x^{24})^2 = (1 - x^6)^7(1 - x^8)^4\tag1$$ $$x(1 - x^{12})^2 = (1 - x^4)^\color{blue}2(1 - x^6)^3\tag2$$
II. Let $n=3$ and $y = \frac1T$ with tribonacci constant $T$. Then,
$$y (1 - y^8)^2 = (1 - y^2)(1 - y^4)^\color{blue}3\tag3$$
III. Let $n=5$ and $z = \frac1P$ with pentanacci constant $P$. Then,
$$z(1 - z^8)^2(1 - z^{12})^4 = (1 - z^2)^2(1 - z^6)^\color{blue}5(1 - z^{24})\tag4$$
And this is where the pattern apparently stops.
The first $(1)$ was mentioned by D. Broadhurst in Multiple Landen values and the tribonacci numbers. (I corrected his typo involving the first factor $x$ of the LHS.) The rest were found by the OP using Mathematica.
Q: Is it possible to find $n=7$?
I expanded my search parameters using $1-q^{m}$ for all even $m\leq32$ but couldn't find anything, nor the next prime $n=7$, so I'm wondering what so special about orders $n=2,3,5$.
P.S. See this post for a related family for all $n$.