Put a term order on your (noncommutative) monomials such that $a_i a_j > a_j a_i$ for $i \lt j$. So the leading term of your equation is $a_1 a_2 a_3$. A basis for your ring is (noncommutative) monomials not divisible by $a_1 a_2 a_3$. In other words, a basis for the degree $n$ part of your algebra is length $n$ sequences of $1$'s, $2$'s and $3$'s which don't contain the sequence $123$. The rest of this post is the combinatorial task of counting the number of such sequences.
Let $A_n$ be the number of such sequences ending in $1$. Let $B_n$ be the number of such sequences ending in $12$. Let $C_n$ be the number of such sequences not in the other two classes (including the empty sequence). Then,
$$A_n = A_{n-1}+ B_{n-1} + C_{n-1}$$ $$B_n = A_{n-1}$$ $$C_n = A_{n-1} + B_{n-1} + 2 C_{n-1} + [n=0]$$
Where $[n=0]$ is $1$ if $n=0$ and is $0$ otherwise. So $$\begin{pmatrix} A_n \\ B_n \\ C_n \\ \end{pmatrix} = \begin{pmatrix} 1 & 1& 1 \\ 1 & 0 & 0 \\ 1 & 1 & 2 \\ \end{pmatrix}^n \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix}$$ The total number of terms of degree $n$ is $A_n+B_n+C_n$, so $$\begin{pmatrix} 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1& 1 \\ 1 & 0 & 0 \\ 1 & 1 & 2 \\ \end{pmatrix}^n \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix}$$
The spectral radius of this matrix is $\approx 2.87939$, so your Hilbert series grow exponentially. That is very different from a commutative ring, whose Hilbert series will grow polynomially.
With a little hacking around with Mathematica, I get that the Hilbert series is $\frac{1}{1-3x+x^3} = 1+3x+9x^2+26x^3+75 x^4 + 216 x^5 + 622 x^6 + \cdots$. Does that match your data?