For any $i,j,k$, the automorphism group of $A$ is transitive on the set of pairs $(I,J)$ such that $|I|=i, |J|=j, |I\cap J|=k$. Therefore the same is true of the inverse (if it exists). That is, the $(I,J)$-th entry of the inverse is $f(i,j,k)$ for some function $f$. I'm too lazy, but I bet that by examining Benjamin's example the function $f(i,j,k)$ can be guessed rather easily. Then we will have an explicit formula for the inverse.

Here's a guess:  $(I,J)$-th entry of the inverse is 0 unless $|I\cup J|=n$ and otherwise is $(-1)^{n+k+1}$.

For any $i,j,k$, the automorphism group of $A$ is transitive on the set of pairs $(I,J)$ such that $|I|=i, |J|=j, |I\cap J|=k$. Therefore the same is true of the inverse (if it exists). That is, the $(I,J)$-th entry of the inverse is $f(i,j,k)$ for some function $f$. I'm too lazy, but I bet that by examining Benjamin's example the function $f(i,j,k)$ can be guessed rather easily. Then we will have an explicit formula for the inverse.

Here's a WRONG guess: The $(I,J)$-th entry of the inverse is 0 unless $|I\cup J|=n$ and otherwise is $(-1)^{n+k+1}$.

Here's a RIGHT guess: The $(I,J)$-th entry of the inverse is 0 if $|I\cup J|\lt n$ and otherwise equals $(-1)^{k+1}$. I checked this up to n=8.

This is easy to prove by induction using Benjamin's recursive formula for the inverse.