= Solution
For a localized charge distribution and a field point $r$ outside it, expand the Coulomb <kernel>:
$$
\frac1{|r-x|}=\frac1r+\frac{r\cdot x}{r^3}+\frac{3(r\cdot x)^2-r^2|x|^2}{2r^5}+O(|x|^3/r^4).
$$
Integrating against the charge density gives the <electrostatic multipole expansion>
$$
\boxed{\phi(r)=\frac1{4\pi\epsilon_0}\left[\frac{Q}{r}+\frac{p_i r_i}{r^3}+\frac{Q_{ij}r_ir_j}{2r^5}+\cdots\right],}
$$
where $Q=\int\rho\,d^3x$, $p_i=\int\rho x_i\,d^3x$ is the <electric dipole moment>, and the symmetric trace-free <electric quadrupole moment> is $Q_{ij}=\int\rho(3x_ix_j-|x|^2\delta_{ij})\,d^3x$. This states the tensor normalization explicitly.
For the tetrahedral point charges, the total charge is zero and the three other <vertex> vectors sum to minus $(1,1,1)$. Hence
$$
\boxed{\mathbf p=4q(1,1,1).}
$$
Every <vertex> has $|x|^2=3$ and each $x_i^2=1$, so each diagonal quadrupole entry vanishes. For each distinct coordinate pair the charge-weighted sum of $x_ix_j$ is $4q$, yielding
$$
\boxed{(Q_{ij})=12q\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}.}
$$
The tensor is trace-free as required. If the alternative raw second-moment convention is used, its off-diagonal entries are $4q$ and the <kernel> contraction has the corresponding factor of three; the physical potential is unchanged.
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