I'm answering my own question (I hope that's within the M.SE etiquette) - I think I've found a partial solution.
Assuming $A$ is nonsingular, the normal vector is given by $v(A) = A^{-1}*1^T$ (I'm not sure why this is true, but I coded a quick simulation to test it and it checks out). We can normalize with $v(A) = \frac{A^{-1}*1^T}{1*A^{-1}*1^T}$, and then apply the standard rules of matrix calculus to get a solution from here:
$$\frac{dv}{dA_{ij}} = (\frac{d(1*A^{-1}*1^T)^{-1}}{dA_{ij}}) A^{-1}*1^T + (1*A^{-1}*1^T)^{-1}(\frac{d(A^{-1}*1^T)}{dA_{ij}})$$
$$ = (-\frac{d(1*A^{-1}*1^T)}{dA_{ij}})^{-2} A^{-1}*1^T + (1*A^{-1}*1^T)^{-1}(\frac{dA^{-1}}{dA_{ij}})*1^T$$
$$ = (-\frac{d(1*A^{-1}*1^T)}{dA_{ij}})^{-2} A^{-1}*1^T + (1*A^{-1}*1^T)^{-1}(\frac{dA^{-1}}{dA_{ij}})*1^T$$
$$ = (-1*\frac{dA^{-1}}{dA_{ij}}*1^T)^{-2} A^{-1}*1^T + (1*A^{-1}*1^T)^{-1}(\frac{dA^{-1}}{dA_{ij}})*1^T$$
$$ = (-1*\frac{dA^{-1}}{dA_{ij}}*1^T)^{-2} A^{-1}*1^T + (1*A^{-1}*1^T)^{-1}(\frac{dA^{-1}}{dA_{ij}})*1^T$$
$$ = (1*A^{-1}*\frac{dA}{dA_{ij}}*A^{-1}*1^T)^{-2} A^{-1}*1^T - (1*A^{-1}*1^T)^{-1}(A^{-1}*\frac{dA}{dA_{ij}}*A^{-1}*1^T)$$
Turns out that $A^{-1}*\frac{dA}{dA_{ij}}*A^{-1}*1^T = (\sum_{y=1}^n A^{-1}_{jy}) (A^{-1}_{*i})$.
$$ = (1*(\sum_{y=1}^n A^{-1}_{jy}) (A^{-1}_{*i}))^{-2} A^{-1}*1^T - (1*A^{-1}*1^T)^{-1}(\sum_{y=1}^n A^{-1}_{jy}) (A^{-1}_{*i})$$
$$ = ((\sum_{y=1}^n A^{-1}_{jy}) (\sum_{x=1}^n A^{-1}_{xi}))^{-2} A^{-1}*1^T - (\sum_{x=1}^n \sum_{y=1}^n A^{-1}_{xy})^{-1}(\sum_{y=1}^n A^{-1}_{jy}) (A^{-1}_{*i})$$
Alright, that's all the simplifying I'm going to do - I really hope my algebra went okay. Hopefully this is helpful to someone else.