Is there some simple upper bound on $(B^{1}+A^{1})^{1}$, where $A,B$ are $n \times n$ symmetric matrices?

You can use the surprising identity $(A^{1}+B^{1})^{1}=A(A+B)^{1}B$, and take the norms of the three factors separately. 


To expand on my first comment, if $A, B > 0$ are symmetric positive definite matrices. Then, it is known that $$\left(\frac{A^{1}+B^{1}}{2}\right)^{1} \le A\sharp B \le \frac{A+B}{2},$$ where the inequalities are in the Löwner partial order, and $A\sharp B := A^{1/2}(A^{1/2}BA^{1/2})^{1/2}A^{1/2}$ denotes the matrix geometric mean. These operator inequalities are of course, stronger than corresponding norm inequalities (based on unitarily invariant norms). For the case where you don't have positive matrices, I think the conjecture mentioned in my second argument can be expanded into a proof  maybe if I get time, I'll try to expand that. 


Probably not unless $A$ and $B$ are positivedefinite, since if $B$ is very close to $A$ then $B^{1}+A^{1}$ is very small and so its inverse is very large. In fact, depending on the norm, they probably need to be close only on one shared or almostshared eigenvector. For spectral norm of positivedefinite matrices, we have a nice answer. The highest eigenvalue of $(A^{1}+B^{1})^{1}$ is the lowest eigenvalue of $A^{1}+B^{1}$, which one can find by minimizing $x^T(A^{1}+B^{1})x$ with respect to $x^Tx=1$. But the minimum for $A^{1}$ is its lowest eigenvalue, $1/A$, and the minimum for $B^{1}$ is similarly $1/B$. Thus: $x^T( A^{1}+B^{1})^{1} x= x^T A^{1} x+ x^T B^{1} x\geq 1/A+1/B$ So the spectral norm of the harmonic sum is bounded by the harmonic sum of the spectral norms! 

