If we're lucky, the $f_j$, $1\le j\le n$ (I use index notation for partial derivatives) are already linearly independent. If not, then $v\cdot\nabla f=0$ for some vector $v\not= 0$, but then also $Pv\cdot\nabla f=0$ for all permutations of the components of $v$. I believe there are two possibilities, though I haven't proved this carefully: the $Pv$ will typically an $(n-1)$ dimensional subspace, so that $f$ can only depend on $w\cdot x$; this corresponds to the exceptional example you mentioned. The only way to avoid such a large span is to have specifically $v=(1,1,\ldots , 1)$.
In this case, we can write $$ f(x_1,x_2,\ldots ,x_n)=f(0,x_2-x_1, \ldots, x_n-x_1) = g(x_2-x_1,\ldots, x_n-x_1) , $$ and $g(t_2,\ldots , t_n)=f(0,t_2,\ldots ,t_n)$ is still a symmetric homogeneous polynomial of its arguments.
Now suppose we have an additional linear relation $\sum a_jf_j=\sum x_j f_{jj}$. We can rewrite this as $$ \sum_{j\ge 2} b_j g_j(t_2,\ldots ,t_n) - \sum_{j\ge 2} t_j g_{jj}(t_2,\ldots ,t_n) = x_1\Delta f(0,t_2,\ldots ,t_n) , $$ with $t_j=x_j-x_1$ and $b_j=a_j-a_1$. Since $x_1$ and the $t_j$'s can be varied independently, both sides must be (identically) equal to zero. Now $g$ is symmetric, so we can as above manipulate for example the first term as follows: $$ \left( \sum b_j D_j\right) g(t_2,\ldots, t_n) = \left( \sum b_j D_j\right) g(t_{\pi 2},\ldots, t_{\pi n}) = \sum b_{\pi k} g_k (t_{\pi 2},\ldots ,t_{\pi n}) $$ The same procedure applied to $\sum t_j g_{jj}$ produces $$ \sum t_{\pi k} g_{kk}(t_{\pi 2}, \ldots , t_{\pi n}) , $$ so by comparing these we obtain that in fact $\sum b_{\pi k} g_k(t) +\sum t_k g_{kk}(t)=0$ for all permutations $\pi$ (this worked because the $b_j$'s, unlike the $t_j$'s, do not get reordered when we relabel the variables at the end). Thus we also have that $$ \sum (b_{\pi k}-b_k) g_k = 0 . \quad\quad\quad\quad (1) $$
We can now run an induction on $n$: if (1) gives us a new linear relation, then this together with the one we already had contradicts the induction hypothesis that there is at most one linear relation among the $g_j, E(g)$. So $b_2=\ldots = b_n$, but then also $a_2=\ldots = a_n$. I can now repeat the whole argument with another variable taking the role of $x_1$ to see that $a_1$ also has the same value (for $n\ge 3$; $n=2$ is easy to do directly; also observe that while $g$ depends on my choice of $x_1$ as the distinguished variable, the $a_j$ don't). This again gives me a contradiction because $\sum f_j=0$.