Let $G:=S_n$ and let $V:=k^n$.
Suppose characteristic of $k$ does not divide $n$.
Define $U:=\{(a_,...,a_n \mid \sum a_i=0\}$
If $\exists$ subspace $U' \subseteq U$ s.t.
$\{e\} \neq U'$ and $gU' \subseteq U'$ $\forall g \in S_n$,
Then $U'=U$.
So $\subseteq$ is clear,
Pick $0 \neq \alpha \in U$, then $\alpha = (a_1,...a_n)$
Since characteristic of $k$ does not divide $n$, we know all entries cannot be equal, furthermore we know all entries cannot be zero and. Where do I go from here to ensure $\alpha \in U'$ ? i.e., $U$ has no nontrivial proper invariant subspaces.
Inside $k^n$ we have two invariant subspaces that are easy to write down: $U$ and $V = \langle (1,1,1, \dots,1) \rangle$. If $p|n$ then $V \subset U$, and if not then they intersect trivially and $U \oplus V = k^n$.
Claim: These are the only two invariant proper subspaces in $k^n$.
Let's prove it by induction on $n$, for $n=2$ it is easy to check directly (with the slight caveat that $U=V$ if we are in characteristic $2$ and $n=2$).
Suppose $W$ is another invariant subspace, and lets take $v \in W$ to be a vector where not all its coordinates are equal (if no such vector exists $W=V$). By using the $S_n$ action to flip two unequal coordinates and subtracting what we get from $v$, we obtain a new non-zero vector $v'\in W$ that has at least one coordinate equal to zero. By permuting the coordinates, we can assume that the last coordinate is zero.
Hence we have found a vector $v' \subset W \cap k^{n-1}$ for the standard embedding of $k^{n-1}$ in $k^n$. This intersection $W \cap k^{n-1}$ is a non-empty $S_{n-1}$ invariant subspace of $k^{n-1}$. Hence by induction we know it must be either the space where the sum of the coordinates is zero, the $1$-dimensional invariant subspace, or all of $k^{n-1}$.
If $W \cap k^{n-1} = k^{n-1}$ then it contains the vector $(1,0, \dots)$, and its $S_n$ orbits clearly span $k^n$ so $W$ must be all of $k^{n}$.
If $W \cap k^{n-1}$ is the space where the sum of the coordinates is zero it contains $(1,-1, 0, \dots, 0)$, and the $S_n$-orbit of that spans $U$. Hence $W$ must contain $U$, so $W = U$ or $W = k^n$ as $U$ has codimension 1.
If $W \cap k^{n-1}$ is the one dimensional subspace spanned by $(1,1,\dots, 1, 0)$. By permuting the coordinates, $(0,1,1\dots 1)$ is also in $W$ and by subtracting $(1, 0,0, \dots -1)$ is in $W$. Again, the $S_n$ orbit of this spans all of $U$ so $W = U$ or $W = k^n$.