The formula that emiliocba seeks seems to be as follows.
Let $\chi_3$ be the Dirichlet character mod $3$.  For $k>0$
write $k = 3^e n$ with $n \equiv \pm 1 \bmod 3$.  Then
the number of representations of $k$ by this quadratic form $A_2^3$ is
$$
s(k) :=
9 (3^{2e+1}-\chi(n)) \phantom. \sum_{d|n} \phantom. \chi(n/d)\phantom. d^2.
$$
I append **gp** code that verifies that this holds for $k \leq 432$.

To prove it in general it will be enough to check that
$$
\varphi := 1 + \sum_{k=1}^\infty \phantom. s(k) q^k
$$
is a modular form of weight $3$ and character $\chi$ for $\Gamma_0(3)$,
and to match a few coefficients with the theta function $\theta_{A_2^3}$.
In principle, it is enough to match only the $q^0$ coefficient:
the dual of $A_2^3$ is isomorphic with the scaling of $A_2^3$ by $1/3$,
so by Poisson summation $\theta_{A_2^3}$ is modular also for
the normalizer $\Gamma_0^+(3)$ of $\Gamma_0(3)$ (generated by $\Gamma_0(3)$ and
the involution $w_3 : \tau \longleftrightarrow -1/3\tau\phantom.$);
and $\Gamma_0^+(3)$ has only one cusp,
and no cusp forms of weight less than $6$ (the weight of
$\eta(\tau)^6 \eta(3\tau)^6$), so the normalized Eisenstein series $\varphi$
is the only candidate for $\theta_{A_2^3}$.


    H = 24
    A2 = sum(m=-H,H,sum(n=-H,H,q^(m^2+m*n+n^2))) + O(q^(3*H^2/4+1));
    L = A2^3;

    chi3(m) = kronecker(m,3)
    {
    s(k, e,n) =
      e = valuation(k,3);
      n = k / 3^e;
      9 * (3^(2*e+1)-chi3(n)) * sumdiv(n, d, chi3(n/d)*d^2)
    }

    L == 1 + sum(k=1,3*H^2/4,s(k)*q^k)