There's something called "color Lie algebras", but I haven'rhaven't been able to fully digest their ideas.
Richard Kerner et. al. have considered a similar problem and solved it by embracing the worldrealm of ternary structures. Is there any hope to stay binary in this brave new world?
Edit concerning explicit expressions for commutators
I was trying to make use of the following realization of $q(2)$-algebra (belonging to the strange series q(n)) consisting of bosonic $B_{ij}=x_i \partial_{x_j}+\psi_i \partial_{\psi_j}$ and fermionic $F_{ij} = x_i \partial_{\psi_j}+\psi_i \partial_{x_j}$, $i,j=1,2$. One can check that these generators indeed form q(2) with signs in commutators naturally appearing according to the degree of variables.
Now, for $\mathbb{Z}_n$ case I'm going to change notation to make the motivation more visible and set $G_{ij}^k \equiv x_i^l \partial_{x_j^{l-k}}$ be generators of degree $k$ in $\mathbb{Z}_n$ commuting variables $x_i^l$ of degree $l$.
I use summation over repeating indices so that, for example, $\mathbb{Z}_3$ case looks like $B_{ij} = x_i \partial_{x_j}+\psi_i \partial_{\psi_j}+c_i \partial_{c_j}$, $F_{ij} = \psi_i \partial_{x_j}+c_i \partial_{\psi_j}+x_i \partial_{c_j}$, $C_{ij} = c_i \partial_{x_j}+x_i \partial_{\psi_j}+\psi_i \partial_{c_j}$.
Setting the commutation rules between variables $x_i^l x_j^k = \tilde{g}(i,l;j,k) x_j^k x_i^l$ I immediately descend to the "non-quantum case" (in the sense described above) and set $\tilde{g}(i,l;j,k)\equiv g(l,k)$.
Then I use the following commutation relations $$ [G_{ij}^r, G_{kl}^p] \equiv G_{ij}^r G_{kl}^p - \alpha(r,p) G_{kl}^p G_{ij}^r$$ with the goal of determining $\alpha(r,s)$ from the condition of absence of second derivatives:
$$ [G_{ij}^r, G_{kl}^p] = x_i^s \partial_{x_j^{s-r}} x_k^q \partial_{x_l^{q-p}} - \alpha(r,p) x_k^q \partial_{x_l^{q-p}} x_i^s \partial_{x_j^{s-r}} = g(r-s,q) g(s,q) g(r-s, p-q) x_k^q x_i^s \partial_{x_l^{q-p}}\partial_{x_j^{s-r}} - \alpha(r,p) g(p-q,s)x_k^q x_i^s \partial_{x_l^{q-p}}\partial_{x_j^{s-r}}$$
so that $\alpha(r,p) = \frac{g(r-s,q) g(s,q) g(r-s, p-q)}{g(p-q,s)}$ for every $s$ and $q$ (which is itself a condition on $g(s,q)$).
However the solution I had in mind while writing this, namely, $g(r,s) = \exp(\frac{2\pi i rs}{n})$ doesn't work here, since $\alpha(r,p) = \exp(\frac{2\pi i}{n}(rp+sq-sp-ps-qs))$ depends on $s$ and $q$. This problem originates in the fact that $x_i^r x_j^s =g(r,s) x_j^s x_i^r$ with $g(r,s)$ taking values in the $n$-th root of $1$ is incompatible with abelian $g(r,s)$, since $g(r,s)g(s,r)=1$, as follows from $x_i^r x_j^s =g(r,s) x_j^s x_i^r =g(r,s) g(s,r) x_i^r x_j^s$.
It's probably not very comprehensible right now, but I'm not sure how to say it the right way.