Let me start with the conclusion. @მამუკა ჯიბლაძე has the correct guess: if $n+1$ has two bits in its binary representation, then the kernel has dimension $2$; otherwise, the kernel is exactly one-dimensional. The only nonzero element in the kernel actually has a nice expression in binary representation. Suppose that $n+1=2^{s_1}+\cdots+2^{s_k}$ where $s_1>\cdots >s_k\geq 0$, then the only nonzero element in the kernel is
$$\sum_{I\subsetneq [k], I\neq \emptyset}\sigma_{\sum_{i\in I}2^{s_i}}.$$
The following is a detailed proof for this claim.
I think the main takeaway would be that it is always convenient to compress a system of identities involving binomial coefficients into one identity involving one single polynomial.
For any $\sum_{i=1}^{n}c_i\sigma_i$, we associate to it a homogeneous polynomial $F(x,y)=\sum_{i=1}^{n}c_ix^iy^{n+1-i}$.
My claim is that the system of equations can be compressed into
$$F(x+z,y)+F(z,y)+F(x,y+z)+F(x,z)=0.$$
(I am not sure if this is why you care about the identity, but if this is then I think it will be better to also mention this.)
To show this, first note that the above identity clearly holds when $z=0$.
Therefore we just need to check the coefficients of $[x^ay^bz^c]$ for $c>0$.
Now note that
$$(x+z)^iy^{n+1-i}+z^iy^{n+1-i}+x^i(y+z)^{n+1-i}+x^iz^{n+1-i}$$
$$=\sum_{0<j<i}\binom{i}{j}x^jy^{n+1-i}z^{i-j}+\sum_{0<j<n+1-i}\binom{n+1-i}{j}x^iy^{n+1-i-j}z^j.$$
Note that the second summand can be written as
$$\sum_{i<j<n+1}\binom{n+1-i}{j-i}x^iy^{n+1-j}z^{j-i},$$
and so if we correspond $\sigma_{i,j}$ to $x^iy^{n+1-i}z^{j-i}$, we see that $\sigma_i$ is indeed mapped to $f_n(\sigma_i)$ via $F(x,y)\mapsto F(x+z,y)+F(z,y)+F(x,y+z)+F(x,z)$.
From now on, we will work with $F$ directly as it exhibits the symmetry better.
Let $I\subseteq\{(1,n),\ldots,(n,1)\}$ and suppose that $F=\sum_{(a,b)\in I}x^ay^b$.
Then by Lucas' theorem,
$$[x^ay^bz^c]F(x+z,y)+F(z,y)+F(x,y)=1$$
if and only if $a,b,c\geq 1$, $a+b+c=n+1$, $(a+c,b)\in I$ and $a+c= a\oplus c$.
The identity $F(x+z,y)+F(z,y)+F(x,y+z)+F(x,z)=0$ is then saying that for $(a,b,c)\in\mathbb{N}^3$ such that $a+b+c=n+1$, we have $(a+c,b)\in I$ and $a+c=a\oplus c$ if and only if $(a,b+c)\in I$ and $b+c=b\oplus c$.
Now we show that if $(a,b)\in I$, then $a+b=a\oplus b$ unless $a=b$ are both a power of $2$.
Otherwise, there exists $c\neq 0$ such that $a\oplus c = a-c$ and $b\oplus c = b-c$ while one of $a-c$ and $b-c$ is nonzero (say, take $c$ to be the number represented by a bit that is $1$ both in $a$ and $b$).
Without loss of generality, assume $a\neq c$.
As $(a,b)\in I$ and $(a\oplus c)\oplus c = (a\oplus c)+c$, we know by the conclusion in the previous paragraph that $b\oplus c=b+c$, which is a contradiction.
Therefore, unless $k=1$, the only candidates of elements in $I$ are those that split the bits of $n+1$ into two parts.
We can now rewrite everything as follows: recall that $n+1=2^{s_1}+\cdots +2^{s_k}$ where $s_1>\cdots>s_k\geq 0$, and set $S=\{s_1,\ldots, s_k\}$.
Then each element in $I$ can be rewritten as a nontrivial bipartition $A\sqcup B=S$.
The conclusion in the previous paragraph says that if $A\sqcup B\sqcup C=S$, then $(A\sqcup C, B)\in I$ if and only if $(A,B\sqcup C)\in I$.
We now deal with the case $k\geq 3$ first.
We first show that if $(A,B)\in I$, then for any $A'\subseteq A$ and $B'\subseteq B'$ that are nonempty, we also have $(A',S\backslash A'), (S\backslash B', B')$ are in $I$.
To see this, simply apply the conclusion above to $(A', B, A\backslash A')$ and $(A, B', B\backslash B')$.
As a consequence, if $I$ is nonempty, then there is some singleton $\{s\}$ such that $(\{s\}, S\backslash \{s\})\in I$, which in turn shows that $(S\backslash \{s'\},\{s'\})\in I$ for all $s'\neq s$.
As $|S|\geq 3$, this shows that $(\{s''\}, S\backslash \{s''\})\in I$ for any $s''\in S$.
As any nontrivial bipartition $S=A\sqcup B$ must satisfy that $B\subseteq S\backslash \{s''\}$ for some $s''$, we get that $I$ must contain all nontrivial bipartition.
Combined with the previous paragraph which says that there are no other possible candidates, we have completely determined $I$, and it is easy to verify that this indeed gives a solution.
The remaining case is $k=1,2$.
When $k=2$, we have already shown that $I\subseteq\{(\{s_1\},\{s_2\}),(\{s_2\},\{s_1\})\}$ as those are the only nontrivial bipartition of $\{s_1,s_2\}$.
It is easy to show that all subsets $I$ work, which corresponds to a $2$-dimensional kernel.
Lastly, when $k=1$, we note that there is no nontrivial bipartition of a singleton set, and thus the only candidate of elements in $I$ is the exceptional case $(a,b)=(2^{s_1-1},2^{s_1-1})$.
It is easy to show that this indeed corresponds to the $1$-dimensional kernel in this case too.
