Let $G$ be a graph embedded (without crossings) on a torus $T$. It's fairly well known that this implies the chromatic number of $G$ is at most 7. If I lift $G$ to the universal cover of $T$, we get a doubly periodic planar graph $\tilde{G}$ and of course the four color theorem tells us there is a four coloring of $\tilde{G}$.
With a little work I can improve this slightly to say that for any such $G$ there is a finite cover $\widehat{T}$ such that the corresponding cover $\widehat{G}$ is four colorable. My question is: Can this be done uniformly in $G$? If so, how small can we take the cover?
Concretely: Does there exist a covering map $T' \to T$ such the pull back to $T'$ of any graph embedded on $T$ can be properly four colored? Which covers work and what is the minimal degree of such a cover?
I was especially interested in the case where $T = \mathbb{R}^2/\mathbb{Z}^2$ and $T'$ was the 4-fold cover $\mathbb{R}^2/(2\mathbb{Z})^2$ but would be interested in hearing about any case.
EDIT: Since I thought this was a fun question I thought about it more and did some more searching through literature. Here are my current best partial results:
For a surface $\Sigma$ of genus $g$ there exists a degree $36^g$ cover such that any graph embedded on $\Sigma$ becomes $6$-colorable when pulled back to the cover.
For genus 1, any graph embedded on a torus becomes $5$-colorable when pulled back to the $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ cover described above.