Let $\mathbb{F}$ be a field of characteristic $2$, $n$ be a positive integer and $f_n:\bigoplus\limits_{i=1}^n\mathbb{F}\sigma_{i}\mapsto \bigoplus\limits_{i,j=1,i<j}^n\mathbb{F}\sigma_{i,j}$ be a linear map, where we identify $\sigma_{i,j}=\sigma_{j,i}~\forall~i,j\in\lbrace 1,\dots,n\rbrace$, which is defined as: 
$$f_n(\sigma_i)=\sum_{j<i}\binom{i}{j}\sigma_{i,j}+\sum_{i<j}\binom{n-i+1}{j-i}\sigma_{i,j}~\forall~i=1,\dots,n$$
I want to compute $\dim_{\mathbb{F}}(\ker(f_n))$. Since we are in characteristic $2$ we would need to know when the combinatorial numbers are even or odd, for which there are theorems like Kummer's theorem, or Lucas' theorem. Sierpinski-Pascal-Triangle give us a geometric description of the parity of the combinatorial numbers, however I don't have ideas on how to use these facts to solve the problem.

Thanks for your help.

Edit: Cheking some small values by hand I got that:
$$\ker(f_3)=\mathbb{F}\sigma_2\Rightarrow\dim(\ker(f_3))=1$$
$$\ker(f_4)=\mathbb{F}\sigma_1\oplus\mathbb{F}\sigma_4\Rightarrow\dim(\ker(f_4))=2$$
$$\ker(f_5)=\mathbb{F}\sigma_2\oplus\mathbb{F}\sigma_4\Rightarrow\dim(\ker(f_5))=2$$
Moreover, it is easy to see that there is some kind of symmetry, for example if $f_n(\sigma_i)=0$ then $f_n(\sigma_{n-i+1})=0$ and in general, if $f_n(\sigma_i)=\sum\limits_{j\neq i} a_j\sigma_{i,j}$ with $a_j\in \lbrace0,1\rbrace$ then $f_n(\sigma_{n-i+1})=\sum\limits_{j\neq i} a_j\sigma_{n-i+1,n-j+1}$

As a first step I would like to know in which cases $f_n(\sigma_i)=0$ and maybe we can prove that those form a basis for $\ker(f_n)$.

Edit: Using the ideas of @მამუკა ჯიბლაძე I proved the following easy facts:

- If $n=2^{m+1}-1$ then $f_n(\sigma_{2^m})=0$
- If $n=2^{m}+2^{k}-1$ then $f_n(\sigma_{2^m})=f_n(\sigma_{2^k})=0$
- If $f_n(\sum_{i=0}^n\lambda_i\sigma_i)=0$ then: if $n$ even $\lambda_i=\lambda_{i+1}~\forall~i$ even and if $n$ odd $\lambda_i=0$ for $i$ odd.

However, I am not able to prove anything else.