Suppose $G$ is a connected semisimple linear algebraic group with Lie algebra $\mathfrak{g}$ and $X$ is a homogeneous $G$-space with isotropy subgroup $H$ (associated Lie algebra $\mathfrak{h}$) that is normalised by a parabolic subgroup $P$. Write $A=P/H$.

Working over $\mathbb{C}$, Borho and Brylinski explain in Proposition 2.8 of "Differential Operators on Homogeneous Spaces I" how to use the moment map $T^\ast X\rightarrow \mathfrak{g}^\ast$ to induce a map from $\pi\colon T^\ast X/A\rightarrow \mathfrak{g}^\ast$. We may also understand this map as a map from the vector bundle $G\times^P\mathfrak{h}^\perp$ to $\mathfrak{g}^\ast$. It is a generalisation of the Springer resolution (which arises in case $H=P$ is a Borel). 

The same construction can be made over an algebraically closed field of characterstic $p>0$. My question is what is known about the dimensions of the fibres of $\pi$ in this characteristic $p$ case? I am most interested in the knowing the largest possible dimension of a fibre over a non-zero point in $\mathfrak{g}^\ast$ in the case where $P$ is a Borel and $H$ is its unipotent radical but more general results where $P$ is any parobolic and $H$ is its unipotent radical are also of interest. I am not interested in the case $H=P$.