Some comments.
Check this question and the comments.
A paper from the answer
Proposition 6. No algorithm is possible that, given a polynomial mapping
$f : \mathbb{R}^n \to \mathbb{R}^n$ with computable coefficients, decides whether this mapping is
surjective.
On the other hand another answer gives relatively efficient criterion
for deciding if the map is bijective over $\mathbb{Q}$ (this implies surjective).
Added
In comments abx asked about existence of surjective quadratic
map for odd $n \ge3$.
Here is an example of an invertible polynomial map for $n=3$.
$$ f_1= {x}^{2}+13\,xy+{y}^{2}+2\,x+y+11\,z+1$$
$$ f_2= xy+x+y+z+1 $$
$$ f_3= xy+z$$
The inverse is:
x = -C^2-12*C+2*C*B-B^2+A-1+B, y = C^2+11*C-2*C*B+B^2-A, z = -A-4*C^3*B+6*C^2*B^2-2*C^2*A-47*C^2*B+25*C*B^2-23*C*A-4*C*B^3-2*B^2*A+B*A+4*C*B*A+B^4+A^2+12*C+B^2-13*C*B-B^3+23*C^3+C^4+133*C^2
