This is more an idea to explore than a complete answer.
You may interpret the binomial coefficient $\binom{n}{k}$ as the elementary symmetric function $e_k$ of $1,1,\ldots,1$ ($n$ variables evaluated at $1$). The coefficients of the adjoint matrix of $A_n$ become skew Schur functions of $1,1,\ldots,1$. Then there may be some further simplifications.
(By the way, this approach gives a nice proof for the value of the determinant of $A_n$: it is the value of the staircase Schur function $s_{(n-1,n-2,\ldots,1,0)}$ evaluated at $1,1,\ldots,1$. Note that the staircase Schur function at $x_1,x_2,\ldots,x_n$ is equal to $\prod_{i \lt j} (x_i+x_j)$).
EDIT: I find that the coefficient $(i,j)$ of the inverse is $(-1)^{i+j} s_{[j]'/(n-i)}(1,1,\ldots,1)/2^{\binom{n}{2}}$, where $[j]$ stands for the partition obtained from $(n-1,n-2,\ldots,1)$ by removing $j$, and $[j]'$ is its conjugate. At this point there is some hope to find a nice formula. First by expressing the skew Schur function as a sum a Schur functions by means of dual Pieri rule: $$ s_{\lambda/(k)}=\sum s_{\nu} $$ where the sum is carried over all partitions $\nu$ obtained from $\lambda$ by removing a horizontal strip with $k$ boxes. After that by using the following formula found in Macdonald, I.3. Ex. 4: $$ s_{\lambda'}(1,1,\ldots,1)=\prod_{x \in \lambda} \frac{n-c(x)}{h(x)} $$ where the evaluation is at $(1,1,\ldots,1)$ with $n$ ones, $\lambda'$ is the conjugate of $\lambda$ and $h(x)$ and $c(x)$ are the hook lengh and content respectively of the box $x$ in the diagram of $\lambda$.
Hopefully the formulas simplify.