Let $n \ge 7$. If $V$ is an irreducible representation of $S_n$ such that $\mathbb{C}[V]^{S_n}$ is a polynomial algebra then either $V$ is the trivial representation, the sign representation or the $(n-1)$-dimensional standard representation. 

<i>Outline Proof</i>: Let $\rho : S_n \rightarrow \mathrm{GL}(V)$ be an irreducible representation affording the irreducible character $\chi^\lambda$ where $\lambda$ is a partition of $n$. Suppose that $V$ is $d$-dimensional. By the Chevalley-Shephard-Todd theorem, $\mathbb{C}[V]^{S_n}$ is a polynomial algebra if and only if $\rho(S_n)$ is generated by pseudo-reflections. If $\rho(g)$ is a pseudo-reflection then $\rho(g)$ is similar to a diagonal matrix $\mathrm{diag}(1,1,\ldots,1,\zeta)$ where $\zeta$ is a root of unity. Hence $\chi(g) = d-1 + \zeta$. However, the irreducible characters of symmetric groups are real valued, so if $g \not= 1_{S_n}$ then $g$ is an involution and $\chi^\lambda(g) = d-2$. 

It therefore suffices to show that if $n \ge 7$ and $g \in S_n$ is an involution such that $\chi^\lambda(g) = \chi^\lambda(1)-2$ then $g$ is a transposition and either $\lambda = (1^n)$ or $\lambda = (n-1,1)$. This follows by induction using the Murnaghan-Nakayama rule. (The details are fiddly but routine.)

For smaller $n$ there are two exceptional cases, corresponding to the partitions $(2,2)$ of $4$ and $(2,2,2)$ of $6$. The former representation is $2$-dimensional and is obtained from the standard representation of $S_3$ via the quotient map $S_4 \rightarrow S_3$. The latter is $5$-dimensional, and can be obtained by applying the outer automorphism of $S_6$ to the standard $5$-dimensional representation of $S_6$; the relevant character value is $\chi^{(2,2,2)}(g) = 3$ where $g = (12)(34)(56) \in S_6$.