Sets that can be mapped onto R^n by a polynomial

The answer to the precise version of the question (the one asked at the end) is no.

The map $(x,y) mapsto (x,xy)$ maps $mathbb{R}^2$ onto all of $mathbb{R}^2$ except the positive and negative parts of the $y$-axis. Following this with $z mapsto z^3$ (where $z=x+iy in mathbb{C}$) yields a polynomial surjection $(x,y) mapsto (f_1(x,y),f_2(x,y))$ such that the preimage of $(0,0)$ is the $y$-axis. Define $f(x,y,z) = (f_1(x,y),f_2(x,y),z)$, so $f$ defines a surjection $mathbb{R}^3 to mathbb{R}^3$. Let $U := mathbb{R}^3 - lbrace (0,y,z) : y le e^z rbrace$, so $U$ is a topological ball. The value of $f$ on each deleted point $(0,y,z)$ is the same as its value at $(0,e^z+1,z) in U$, so $f|_U$ is still surjective.

For $r>0$, if $u in U$ and $f(u) = (0,0,log r)$, then $u=(0,y,log r)$ with $y>r$, so $|u|>r$. So $f(U cap B(r))$ does not even contain $(0,0,log r)$, let alone $B(r^epsilon)$ for large $r$ (for fixed $epsilon>0$).