I'm afraid the marginal distribution for the influence of a single coordinate is the easier part :-) I gave up on this problem because the interrelations between the influences of different coordinates seem too complicated. But if you're interested in the marginal distribution, here's why it's binomial: A given influence for the $i$-th coordinate corresponds to a given number $m$ of pairs of points differing only in the $i$-th coordinate that have different function values. There are $2^{k-1}\choose m$ different ways of choosing the remaining $k-1$ coordinates for these $m$ pairs. For each pair with different function values, there are $2$ possibilities, $(-1,1)$ and $(1,-1)$, and for each pair with the same function values, there are also $2$ possibilities, $(-1,-1)$ and $(1,1)$. Thus the factor $2^{2^{k-1}}$ arising from these possibilities is independent of $m$ and cancels out in the probabilities, which are therefore completely determined by the number $2^{k-1}\choose m$ of choices for the coordinates for the pairs.
Here's another idea I tried: We can write the influence of the $i$-th coordinate as
$\operatorname{Inf}_i(f)=\Pr_{x}[f(x)\neq f(\hat x_i)]=\frac{1-2^{-n}\sum_xf(x)f(\hat x_i)}2\;.$
With the Walsh–Fourier transform of $f$,
$f(x)=\sum_j a_jf_j(x)\;,$
this becomes
\begin{align} \operatorname{Inf}_i(f) &=\frac{1-2^{-n}\sum_x(\sum_j a_jf_j(x))(\sum_{j'} a_{j'}f_{j'}(\hat x_i))}2\\ &=\frac{1-2^{-n}\sum_j\sum_{j'}a_ja_{j'}\sum_xf_j(x)f_{j'}(\hat x_i)}2\\ &=\frac{1-2^{-n}\sum_j\sum_{j'}a_ja_{j'}\sum_xf_j(x)(\pm f_{j'}(x))}2\\ &=\frac{1-\sum_j(\pm a_j^2)}2\;,\\ \end{align}
where the sign depends on whether $f_j$ changes sign with $x_i$.
Unfortunately I think this doesn't get us much closer to solving the problem, because although the $a_j$ are linearly uncorrelated, they're not independent and their squares are highly correlated.