Proof that Coxeter arrangements obey this identity: Group together summands according to the two-plane spanned by $n_i$ and $n_j$. For any two-plane $H$, every summand coming from that two-plane is divisible by $\prod_{n_k \not \in H} \langle n_k, v \rangle$. Factoring out this common summand, the contribution from $H$ is $$\sum_{n_i \neq n_j,\ n_i, n_j \in H} \langle n_i, n_j \rangle \prod_{n_k \in H,\ n_k \neq n_i, n_j} \langle n_k, v \rangle.$$
This is the two dimensional example you've already done.
Proof that only Coxeter arrangements obey this identity: Consider $H$, a two plane spanned by some $(n_i, n_j)$. Our first goal is to show that $H \cap \{ n_k \}$ is a dihedral root system.
Let $r$ be the number of hyperplanes in your arrangement. Let $S$ be the ring of polynomial functions and let $I$ be the ideal generated by the functions $\langle n, \ \rangle$, for $n \in H$. Note that every term of your sum which does not come from $(n_i, n_j)$ with $n_i$, $n_j \in H$ lies in $I^{r-1}$. So, the sum of the terms with $n_i$, $n_j \in H$ must be zero modulo $I^{r-1}$.
As before, all of those terms are divisible by $\prod_{n_k \not \in H} \langle n_k, v \rangle$. This is not a zero divisor in $\mathbb{R}[x]/I^{r-1}$$S/I^{r-1}$. So we can factor it out and deduce that $$\sum_{n_i \neq n_j,\ n_i, n_j \in H} \langle n_i, n_j \rangle \prod_{n_k \in H,\ n_k \neq n_i, n_j} \langle n_k, v \rangle \equiv 0 \ \mathrm{mod} \ I^{r-1}$$ But the left hand side is degree $r-2$, so it must be identically zero. By the two dimensional example which you have already done, this shows that $\{ n_k: n_k \in H \}$ is a root system.
So, for any $n_i$ and $n_j$, the set of $n_k$ in the span of $(n_i, n_j)$ is a root system. In particular, the reflection of $n_i$ by $n_j$ is some $n_k$. So your whole set of vectors is a root system.
Warning: I have not, myself, checked the two dimensional case which I am relying on.