Is $\mathbb Z[\sqrt{-3}]$ Euclidean under some other norm?

It isn't possible. If it were, then $\mathbb{Z}[\sqrt{-3}]$ would be a Unique Factorization Domain. But $$4=(2)(2)=(1-\sqrt{-3})(1+\sqrt{-3}),$$ and $2$ and $1\pm\sqrt{-3}$ are non-associate irreducibles.

Alternately, $2$ is irreducible in our ring. But $2$ is not prime, since $2$ divides the product $(1-\sqrt{-3})(1+\sqrt{-3})$, but $2$ divides neither $1-\sqrt{-3}$ nor $1+\sqrt{-3}$.