So you have a cyclic group acting on complex affine space, and the situation at the level of function fields is clear enough: by basic Galois theory the extension of rational functions is a cyclic extension of degree n. It is a Kummer extension, even: write down a linear combination of the indeterminates with nth roots of unity as coefficients in a kind of obvious way, and you have an element whose nth power is cyclic, but transforms under a non-trivial character of the cyclic Galois group. 

I think what you want is a description of the quotient as an affine variety. (Have to go now.)

**Edit:** My thought, given the other answer here, is that the quotient isn't generally an affine space. By the Noether normalization lemma, however, it is not so far from an affine space, though: a finite branched cover, let's say. The Noether lemma is not so "effective" - the usual proof doesn't tell you how to find the polynomial subring to work with. But with the high degree of structure here, perhaps this case is not the worst case. 

**Edit**: Let's try the case n = 3 and think some about $L_1 = z_1+\zeta z_2+ \zeta^{2}z_3$ (in the context of $L_0 = z_1+z_2+z_3$ and $L_2 = z_1+\zeta ^{2} z_2+ \zeta z_3$) where $\zeta$ is a primitive cube root of 1. A cyclic shift one way multiplies $L_1$ by $\zeta$. Therefore $L_1 ^{3}$ is a cyclic polynomial in your sense. Same is true for $L_2 ^{3}$. Since $L_0$ is cyclic also, we seem to be making progress, because the forms $L_i$ are linearly independent (it's a Vandermonde determinant). Hence we have three cyclic polynomials, and we would be close to solving the problem if it were not for the fact that extracting a cube root gives one of three possible answers, for complex solutions. If we knew the cube root to take, it would all come down to Gaussian elimination.