For any odd $ q $ there is a $ [[3,1,2]]_q $ stabilizer code with stabilizer generators $$ XXX,ZZZ^{-2} $$ (for even $ q $ you can define the same code but it will have distance $ d=1 $ because the weight $ 1 $ error $ IIX^{q/2} $ commutes with both stabilizer generators). A $ [[3,1,2]]_q $ code can be purified to a $ [[4,0,3]]_q $ stabilizer state.
For the $ [[3,1,2]]_q $ code given above what are the stabilizer generators of the resulting $ [[4,0,3]]_q $ state for general odd $ q $?
All credit to user2533488 for the answer https://quantumcomputing.stackexchange.com/a/39757/19675 and the accompanying comment.
Let $ \omega $ be a $ q $ root of unity. Then recall that the Pauli matrices for a qudit of size $ q $ have the commutation relation $ X^i Z^j= \omega^{-ij} Z^j X^i $ equivalently $ Z^j X^i = \omega^{ij} Z^j X^i $
For odd $ q $ the stabilizer generators
$XXX$
$ZZZ^{-2}$
yield a $ [[3,1,2]]_q $ stabilizer code. This code has normalizer generators
$XX^{-1}I$
$ZZ^{-1}I$
These normalizer generators have the commutation relation $$ (XX^{-1}I)(ZZ^{-1}I)=\omega^{-2}(ZZ^{-1}I)(XX^{-1}I) $$ and thus we can take $ XX^{-1}I $ to be logical $ X $ and $ ZZ^{-1}I $ to be logical $ Z^2 $. Let $ \{ |\bar{i}\rangle \}_{i=0}^{q-1} $ denote a basis of logical codewords, in other words these are chosen to be eigenvectors of what we have chosen to call the $ Z $ type logical operators in the normalizer. So we have $$ \bar{Z}^2|\bar{i}\rangle = \omega^{2i} |\bar{i}\rangle $$ and $$ \bar{X}|\bar{i}\rangle = |\overline{i+1}\rangle $$ where $ \bar{Z}^2= ZZ^{-1}I $ is logical $ Z^2 $ and $ \bar{X} = (XX^{-1}I)$ is logical $ X $. Here addition is taken modulo $ q $.
Recall that the purification of a single qudit code is always defined to be the state $$ |\psi \rangle= \sum_{i=0}^{q-1} |\bar{i}\rangle |i\rangle $$ and moreover this code will have distance $ d+1 $ if the original code was non-degenerate.
Observe that \begin{align} (\bar{Z}^2 Z^{-2})|\psi \rangle &= (\bar{Z}^2 Z^{-2}) \sum_{i=0}^{q-1} |\bar{i}\rangle |i\rangle \\ &= \sum_{i=0}^{q-1} (\bar{Z}^2|\bar{i}\rangle) (Z^{-2}|i\rangle) \\ &= \sum_{i=0}^{q-1} (\omega^2|\bar{i}\rangle) (\omega^{-2}|i\rangle) \\ &= \sum_{i=0}^{q-1} |\bar{i}\rangle |i\rangle \\ &= |\psi \rangle \\ \end{align} And \begin{align} (\bar{X} X)|\psi \rangle &= (\bar{X} X) \sum_{i=0}^{q-1} |\bar{i}\rangle |i\rangle \\ &= \sum_{i=0}^{q-1} (\bar{X}|\bar{i}\rangle) (X|i\rangle) \\ &= \sum_{i=0}^{q-1} |\overline{i+1}\rangle |i+1\rangle \\ &= \sum_{i=0}^{q-1} |\bar{i}\rangle |i\rangle \\ &= |\psi \rangle \\ \end{align} Again recall that addition is modulo $ q $. And of course we have $ (XXXI)|\psi \rangle=|\psi \rangle $ and $ (ZZZ^{-2}I)|\psi \rangle=|\psi \rangle $ since $ XXX $ and $ ZZZ^{-2} $ stabilize all the $ |\bar{i}\rangle $.
So $ |\psi \rangle $ is a $ [[4,0,3]]_q $ stabilizer state with stabilizer generators \begin{align} &XXXI \\ &ZZZ^{-2}I \\ &XX^{-1}IX \\ &ZZ^{-1}IZ^{-2} \\ \end{align} Note that a $ [[4,0,3]]_q $ is sometimes called an absolutely maximally entangled state of 4 qudits of size $ q $ (see for example Does a 3 qudit code always exist? (except for qubits) and Uniqueness, absolutely maximally entangled states, and the 3 qutrit code).
Also note that for the case of qutrits ( $ q=3 $ ) this all reduces to the well known 3 qutrit code and the corresponding 4 qutrit AME state with stabilizers $ XXXI,ZZZI,XX^{-1}IX,ZZ^{-1}IZ $, see appendix A.2 of Holographic quantum error-correcting codes: Toy models for the bulk/boundary correspondence