Well, your example $R=\Bbb C[x_1,x_2,...,x_n]/(x_1^2-1,x_2^2-1,...,x_n^2-1)$ is a very easy to understand ring: it is just isomorphic to $\mathbb{C}^{2^n}$. Explicitly, given a function $e:\{1,\dots,n\}\to \{-1,1\}$, define a homomorphism $R\to\mathbb{C}$ by sending $x_k$ to $e(k)$ for each $k$. This gives $2^n$ homomorphisms $R\to\mathbb{C}$ which together give a homomorphism $R\to\mathbb{C}^{2^n}$ which is an isomorphism. (There are many ways you can show it is an isomorphism--for instance, you can show it is surjective, and then conclude it must be injective as well since $R$ and $\mathbb{C}^{2^n}$ have the same finite dimension as $\mathbb{C}$-vector spaces.)
Assuming that you say $f$ "factors" if you can write $f=gh$ where neither $g$ nor $h$ is a unit, an element $f\in\mathbb{C}^{2^n}$ factors iff at least one of its coordinates is $0$. Indeed, if the $e$th coordinate of $f$ is $0$, then $f$ factors as $fh$ where $h$ is $1$ in every coordinate except the $e$th and $0$ in the $e$th coordinate. Conversely, if all the coordinates of $f$ are nonzero, $f$ is a unit, so $f=gh$ implies $g$ and $h$ are both units.
Explicitly in terms of polynomials, this means that $f\in R$ factors iff $f(\pm1,\pm1,\dots,\pm1)=0$ for some choice of signs.
If by "factors" you mean you can write $f=gh$ where neither $g$ nor $h$ is a unit multiple of $f$, then $f\in\mathbb{C}^{2^n}$ factors iff at least two of its coordinates are $0$. Indeed, if the $e$ and $e'$ coordinates are $0$, let $g$ be $f$ modified so that its $e$ coordinate is $1$, and let $h$ be $0$ on its $e'$ coordinate and $1$ on all others. Conversely, if $f$ has at most one nonzero coordinate and $f=gh$, then one of $g$ and $h$ must have the same set of nonzero coordinates as $f$, and thus be a unit multiple of $f$.