There is another way to see this than constructing an explicit resolution. This involves viewing $R$ as an iterated skew polynomial ring.
I am assuming that you want $a_{ii} = 1$; otherwise $x_i^2 = 0$ for all $i$.
Also, since this works over any field, I am just going to denote the base field by $k$.

Start with $R_1 = k[x_1]$. Then let $\sigma_1$ be the $k$-algebra automorphism of $R_1$ defined by $\sigma_1(x_1) = a_{21} x_1$, and let $R_2$ be the skew-polynomial ring
$$
R_2 = R_1[x_2; \sigma_1].
$$
Thus $R_2$ is generated by $x_1$ and $x_2$ with the relation
$$
x_2 x_1 = \sigma_1(x_1)x_2 = a_{21} x_1 x_2.
$$
Then we continue this game. Having constructed $R_i$, define
$$
R_{i+1} = R_i[x_{i+1}, \sigma_i],
$$
where $\sigma_i \in \mathrm{Aut}(R_i)$ is defined by $\sigma_i(x_j) = a_{i+1,j} x_j$ for $1 \le j \le i$. Then for $j < i$ we have the relations
$$
x_i x_j = \sigma_{i-1}(x_j) x_i = a_{ij} x_j x_i,
$$
and hence your ring $R$ coincides with $R_n$.

This gives us two things. First, there is an analogue of the Hilbert Basis Theorem for skew polynomial rings; if $A$ is left Noetherian then the skew polynomial ring $A[x;\sigma]$ is left Noetherian for any automorphism $\sigma$ of $A$. You can find this in Section 1.2.9 of McConnell-Robson or Theorem 1.14 of Goodearl-Warfield.

The other fact is that there is an analogue of the (generalized) Hilbert Syzygy Theorem for skew polynomial rings over Noetherian rings. This is in Section 7.9.10 of McConnell-Robson. Explicitly, it says the following: if $A$ is left Noetherian with $\mathrm{l.gl.dim} \, A = n < \infty$, then $\mathrm{l.gl.dim} \, A[x;\sigma] = n+1$ for any automorphism $\sigma$ of $A$.

Starting with $\mathrm{l.gl.dim}k[x] = 1$ and iterating shows that each $R_i$ is both left and right Noetherian and has
$$
\mathrm{l.gl.dim} R_i = \mathrm{r.gl.dim} R_i = i.
$$

Lectures on Modules and Rings. In particular, Section 5B gives methods of computing the right global dimension of $R/I$ when $I$ is nice. I don't know much about the application you have in mind, but it might fit into Lam's framework. Are you aware of the ways to bound right global dimension via regular sequences? That might help, since you know how the ideal $I$ is defined. Lam's book is really a master-piece; I can't recommend it highly enough! $\endgroup$ – David White Aug 1 '12 at 2:50