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I only concern myself with faithful representations of $S_n$ and for $n >4$. The only way to get a polynomial algebra of invariants is to represent $S_n$ as a complex reflection group (so generated by pseudo-reflections, that is elements wih a fixed-point space of codimension $1$). A complex relfection group is easily checked to always be a direct product of ireducible complex reflection groups, so from now on I consider only irreducible representations. The question then becomes: is there an irreducible representation of $S_n$ other than the usual $n-1$-dimensional irreducible constituent of the natural permutation character, where $S_n$ is represented as a complex reflection group? Note that tensoring with the sign representation of the stated $n-1$-dimensional representation does not produce a representation of $S_n$ as a complex reflection group. There is still some content to this question. It can probably be easily addressed by the Murnaghan-Nakayama rule, or from a more detailed knowledge of the character table of $S_n,$ but I attempt a direct argument here. Since the characters of $S_n$ are rational-valued, any element represented as a pseudo reflection in the given representation must be represented as a genuine (or "real") reflection, since its trace must be rational. The generating reflections for $S_n$ in the given representation act with determinant $-1,$ so lie outside the derived group $A_n.$ Hence they are odd permutations, and expressible as a product of an odd number of (disjoint) transpositions, since they have order $2$. Since $A_n$ is simple ( as $n \geq 5$,) any single conjugacy class of these reflections generates the whole of $S_n,$ so we assume that all the generating reflections are conjugate. Note that we are not yet entitled to assume that the generating reflections are transpositions. Now we note that the product of two generating reflections has order $1,2,3,4$ or $6.$ For such a product has a fixed point space of codimension $0$ or $2,$ has determinant $1$, and has a rational trace. Now if $k >1$ is odd, a product of $k$ transpositions inverts a $2k+1$-cycle in $S_{2k+1}.$ Hence if $k > 3,$ and $n \geq 2k+1,$ then $S_{n}$ contains a pair of conjugate permutations, each a product of $k$ disjoint transpositions, whose product has order $2k+1.$ hen $n = 2k$ we may express a $2k$-cycle as a product of two permutations, one a product of $k$ disjoint transpositions, the other a product of $k-1$ disjoint transpositions. Since $k > 3,$ if we adjoin a transposition interchanging the fixed points of the second permutation, we get a product of two permutations, each a product of $k$ disjoint transpositions, where the product is a product of two disjoint cycles of lengths (allowing a $1$-cycle) whose sum adds to $n.$ Since $n \geq 10,$ sucha product never has order $1,2,3,4,5$ or $6.$ Hence we can assume that our generating reflections are products of at most three transpositions. Furthermore, if $n \geq 8,$ we can express a $5$-cycle in $S_n$ as a product of permutations, each a product of three disjoint transpositions. For we may express a $5$-cycle in $S_5$ as a product of two permutatons, each a product of two disjoint two-cycles. Affix a transposition commuting with the $5$-cycle to each of them, and this does not change the product. Hence if our reflections are products of $3$ disjoint $2$-cycles, we are left with the case $n =6.$ In fact, this case does occur. "Twist" the natural (non-unimodular) irreducible $5$-dimensional representation of $S_6$ by the exceptional outer automorphism of $S_6,$ and we obtain a reflection representation of $S_6$ in which products of $3$ disjoint $2$-cycles act as reflections. It remains to deal with reflection representations of $S_n$ in which transposition act as reflections. I presume that the Murnaghn-Nakayama rule or the character table of $S_n$ then shows that only the expected representation arises, but I leave that issue open here. (Later Edit: This now seems to be covered by Mark Wildon's answer).