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Consider the following map $f$ from $\mathbb R^2$ to the torus $\mathbb T^2=\mathbb R^2/\mathbb Z^2$: $$ f(x,y) = (x, \sqrt2 x) \bmod \mathbb Z^2 . $$ Its set of critical values equals its image and it is dense in the torus, so the $\epsilon$-neighborhood has full measure for any $\epsilon>0$.

EDIT: This answer is wrong: as pointed out by Alfonz, this map is not algebraic. The similar one with a rational approximation of $\sqrt2$ is, but then the degree is unbounded.

Consider the following map $f$ from $\mathbb R^2$ to the torus $\mathbb T^2=\mathbb R^2/\mathbb Z^2$: $$ f(x,y) = (x, \sqrt2 x) \bmod \mathbb Z^2 . $$ Its set of critical values equals its image and it is dense in the torus, so the $\epsilon$-neighborhood has full measure for any $\epsilon>0$.

Consider the following map $f$ from $\mathbb R^2$ to the torus $\mathbb T^2=\mathbb R^2/\mathbb Z^2$: $$ f(x,y) = (x, \sqrt2 y) \bmod \mathbb Z^2 . $$ Its set of critical values equals its image and it is dense in the torus, so the $\epsilon$-neighborhood has full measure for any $\epsilon>0$.

Consider the following map $f$ from $\mathbb R^2$ to the torus $\mathbb T^2=\mathbb R^2/\mathbb Z^2$: $$ f(x,y) = (x, \sqrt2 x) \bmod \mathbb Z^2 . $$ Its set of critical values equals its image and it is dense in the torus, so the $\epsilon$-neighborhood has full measure for any $\epsilon>0$.

Consider the following map $f$ from $\mathbb R^2$ to the torus $\mathbb T^2=\mathbb R^2/\mathbb Z^2$: $$ f(x,y) = (x, \sqrt2 y) \bmod \mathbb Z^2 . $$ Its set of critical points equals its image and it is dense in the torus, so the $\epsilon$-neighborhood has full measure for any $\epsilon>0$.

Consider the following map $f$ from $\mathbb R^2$ to the torus $\mathbb T^2=\mathbb R^2/\mathbb Z^2$: $$ f(x,y) = (x, \sqrt2 y) \bmod \mathbb Z^2 . $$ Its set of critical values equals its image and it is dense in the torus, so the $\epsilon$-neighborhood has full measure for any $\epsilon>0$.