Nonnegative Ricci curvature, nilpotency, and Hausdorff dimension

Let $M$ be an open (complete and non-compact) manifold with $\mathrm{Ric}\ge 0$ and escape rate not $1/2$. It is known that under these conditions, the fundamental group $\pi_1(M)$ has a finitely generated torsion-free nilpotent subgroup $\mathcal{N}$ of finite index, as long as $\pi_1(M)$ is an infinite group. We show that the nilpotency step of $\mathcal{N}$ must be reflected in the asymptotic geometry of the universal cover $\widetilde{M}$, in terms of the Hausdorff dimension of an isometric $\mathbb{R}$-orbit: there exist an asymptotic cone $(Y,y)$ of $\widetilde{M}$ and a closed $\mathbb{R}$-subgroup $L$ of the isometry group of $Y$ such that its orbit $Ly$ has Hausdorff dimension at least the nilpotency step of $\mathcal{N}$. This resolves a question raised by Wei and the author.