Let $N$ be a nilpotent group (I'm interested in real linear Lie groups primarily), I'm secretly thinking of $N$ being embedded inside some $G$ such that $N$ is horospherical subgroup (meaning there exists some $a$ for which $a^{-n}Na^{n} \to e$ as $n\to\infty$).
Given such $a$ (which we may assume have a semisimple action on Lie(N)), I define the following sets $B^{N}_{n}=\exp(a_{\log n}B_{1}a_{-\log n})$, where $B_{1} \subset Lie(N)$ is the logarithm of the Riemannian (say left-invariant) metric ball in $N$. What is the volume of $B^{N}_{n}$?
Obviously it has to do with the normalization of the one-param group $a_{\star}$ so imagine that the sum of eigenvalues of $a_{1}$ on $Lie(N)$ equals to $1$. Notice also that when $N$ is a nice nilpotent group (say Heisenberg) then those "rectangular balls" equals to the balls according to the CC metric.
Notice that it is well known that metric balls $B_{N}$ has exact polynomial growth with the exponent given by the Bass-Guivarch formula.
Edit - it would be helpful for me even just to get a precise power growth for this sequence.