Here to determine the number of solutions to the diagonal equation, we need to find number of codewords of weight $i$, i.e. $A_i$, in a formula. But here to calculate those $A_i$, I found no formula for the weight distribution of the linear cyclic codes. In this case the principal ideal ring over which the code is formed is $\mathbb{F}_2[x]/\langle x^9-1\rangle$. But how to calculate the $A_i$'s I don't get here. Any help is appreciated.
$A_0=1$ is obvious. The rest is due to the special structure of $G$.
Clearly no linear combination of the rows of $G$ will yield a weight 1 codeword.
Since the rows $g_i$ of $G$ have weight 2, there are 6 codewords of weight 2. Also note that $g_i\oplus g_{i+3}$ has weight 2 due to cancellation, and there are 3 choices of such $i$, namely $i=1,2,3.$ This means $A_2=9.$
Clearly one cannot get weight 3 by linear combinationrs of the rows of $G$ either.
As for $A_4,$ take all pairs of rows $g_i,g_j$ subject to $i\neq j \pmod 3.$
No codeword of weight 5 can be produced from linear combinations of these rows.
As for $A_6$ I didn't check carefully but taking linear combinations of all triples of rows $g_i,g_j,g_k$ subject to $$i\neq j \pmod 3\quad i \neq k \pmod 3 \quad j\neq k \pmod 3$$ should work.
Then observe $1+9+27+27=64=2^6,$ so there are no more codewords by dimension arguments.