Rita's answer is perfectly correct. However, Jérémy posed his question over a (not necessarily algebrically closed) characteristic $0$ field
$k=\mathbb{Q}$, so here is another approach. Consider the subring $S = k[s_1,\dots,s_n]$ generated by the elementary symmetric polynomials, i.e., the invariant ring under the entire symmetric group, not just the cyclic group. The entire polynomial ring $R = k[x_1,\dots,x_n]$ is an $S$-algebra that is, in fact, a
free $S$-module of rank $n!$. The fact that this is a free $S$-module follows automatically since $S$ is regular and $R$ is Cohen-Macaulay; however, there are also perfectly explicit choices of $S$-basis, e.g.,
the monomials $\underline{x}^{\underline{e}} =x_1^{e_1}x_2^{e_2}\cdots x_n^{e_n}$ indexed by all exponents $(e_1,\dots,e_n)\in \mathbb{Z}_{\geq 0}^n$ with $e_i < i$ for every $i$. The $k$-algebra homomorphism $$\phi:R\to R, \ \ \phi(x_1,\dots,x_{n-1},x_n) = (x_2,\dots,x_n,x_1),$$ is an $S$-algebra homomorphism. The ring of $\phi$-invariants is the kernel of the $S$-module homomorphism $\text{Id}_R-\phi:R\to R$, which is the same as the image of the "averaging homomorphism" / Maschke homomorphism / Reynolds operator $$E=\frac{1}{n} ( \text{Id} + \phi + \phi^2 + \dots + \phi^{n-1} ).$$
In particular, the subring of $\phi$-invariants is a direct summand of $R$ as an $S$-module. Of course one set of generators is the set of images $E(\underline{x}^{\underline{e}})$ as above;
you can compute a smaller set of $S$-generators for the kernel using a term order on $R$ as an $S$-module and using Gröbner basis methods. Any collection of $S$-generators give $k$-algebra generators once you add in the $k$-algebra generators of $S$, i.e., the elementary symmetric polynomials.

Regarding the question when the ring of invariants is a polynomial ring, or even a smooth $k$-algebra, this fails for all $n\geq 3$. As explained in the comments, this can be checked after base change to $\overline{k}$, where it becomes an immediate corollary of the Chevalley-Shephard-Todd theorem (or just direct computation). However, as you can see, the invariant ring is Cohen-Macaulay; it is a free $S$-module (special case of the Hochster-Roberts Theorem).

$\textbf{Edit}$. It seems quite plausible that, as a free $S$-module, the ring of invariants has a basis consisting of $E(\underline{x}^{\underline{e}})$ for those $(n-1)!$ exponents as above satisfying the additional condition that $e_n=0$.