Let $K$ be a field. Consider the vector space $K^n$ over the field $K$. Suppose $(a_1,a_2, ... ,a_n) \in K^n$. What is the dimension of the subspace generated by all the permutations of $(a_1,a_2,...,a_n)$?
I think there are 4 different cases
$a_1=a_2=...=a_n=0$
$a_1=a_2=a_3=...=a_n \ne 0$
$a_1+a_2+...+a_n=0,$ $ a_1 \ne a_2$
$a_1+a_2+...+a_n \ne 0$ , $a_1 \ne a_2$
Let $V$ be the subspace spanned by the permuted versions of the vector $a=(a_1,a_2, ... ,a_n) \in K^n$. Let $U$ be the complementary subspace w.r.t. the usual bilinear form $\langle x,y\rangle=\sum_{i=1}^nx_iy_i$ of $K^n$, i.e. $$ U=\{ x\in K^n\mid \langle x,y\rangle=0\;\text{for all $y\in V$}\} $$ Then it is impossible for both $V$ and $U$ to have a vector with non-equal components. Assume contrariwise that $a_i\neq a_j$ for some pair of indices $i<j$, and that there exists $u=(u_1,u_2,\ldots,u_n)\in U$ such that $u_k\neq u_\ell$ for some pair of indices $k<\ell$. Then we can find two permuted versions of $a$: one, call it $v$, where $a_i$ occurs at position $k$ and $a_j$ at position $\ell$, and another, call it $w$, gotten from $v$ by swapping those two entries. Then $$ 0=0-0=\langle u,v\rangle-\langle u,w\rangle=\langle u,v-w\rangle=(u_k-u_\ell)(a_i-a_j), $$ which is a contradiction.
The conclusion is that one of the subspaces $U$ or $V$ is of dimension at most 1. As they are complementary $\dim U +\dim V=n$, and we can deduce that the dimension of $V$ is either 0,1, $n-1$ or $n$. It is clear, when either of the first two cases occurs. If $\dim V=n-1$, then $\dim U=1$, and, by the above observation, $U$ is spanned by the all ones vector, so $V$ is the zero-sum subspace. When none of these cases applies, we must have $\dim V=n$.
Given the question only asserted that $K$ is ``a field'', the argument above is somewhat problematic: if $K=\mathbb C$, then if $U$ is a subspace of $\mathbb C^n$, the subspace
$$ V = \{w \in K^m: \langle v,w \rangle =0, \forall u \in U \} $$
will not in general give a complementary subspace to $U$. For example, if $n=2$, then if $U=\mathbb C.(1,i)$, the subspace $V$ is equal to $U$. It works over a field like $\mathbb R$, and if you use a Hermitian form instead $\langle x,y \rangle = \sum_{i=1}^n x_i\bar{y_i}$, then it works for $\mathbb C$.
But actually the form isn't really necessary for the heart of the argument above: if $U$ is a subspace of $K^n$ not contained in the line $L = K.(e_1+\ldots+e_n)$ and $U$ is stable under the action of $S_n$, then we claim it is either $K^n$ or
$$ H = \{\sum_{i=1}^n a_ie_i: \sum_{i=1}^n a_i = 0\}$$
To see this, note that if $v=\sum_{i=1}^n c_ie_i \in U$ is not on the line $L$, then there must exist $i<j$ with $c_i\neq c_j$. Then if $\sigma_{ij}$ denotes the permutation which swaps $i$ and $j$ and fixes everything else, we have
$$ v-\sigma_{ij}(v) = c_ie_i+c_je_j -(c_je_i +c_ie_j) = (c_i-c_j)(e_i-e_j) \in U. $$ Thus $e_i-e_j \in U$, and as $U$ is stable under the $S_n$-action, it follows that $U$ contains $e_k-e_l$ for any pair basis vectors, and hence contains $H$ (since, for example, $\{e_1-e_n,e_2-e_n,\ldots,e_{n-1}-e_n\}$ is a basis of $H$). Thus $U = K^n$ or $U=H$ as claimed since $\dim(H)=n-1$.
It follows that the only $S_n$-stable subspaces of $K^n$ are $\{0\},L, H$ and $K^n$ itself. In other words, if you start with a vector $v$ in $K^n$, then the subspace generated by all the permutations of $v$ will be $\{0\}$ if $v=0$, $L$ if $v$ has all of its coordinate equal (that is, if $v \in L$), $H$ if the coordinates of $v$ sum to $0$, and all of $K^n$ otherwise.
It is also worth noting that if the characteristic of $K$ is coprime to $n$, that is, if $n$ is a unit in $K$, then $K^n = H \oplus L$, but if $\text{char}(K)=p$ and $p\mid n$, then $L\leq H<K^n$, so that $K^n$ is not a direct sum of irreducible $S_n$-representations.
