$\mathbb{R}^n$ is not a polyhedron. You mentioned that a generalized halfspace can be all of $\mathbb{R}^n$, but that is not that case. By definition, a halfspace can only emcompass half of $\mathbb{R}^n$. I'm not sure what definition you're given for a generalized halfspace, but it's just the direct analogue of the three dimensional case.
Since a polyhedron is by definition the intersection of a finite collection of halfspaces, if $\mathbb{R}^n$ is a polyhedron then there must be at least one halfspace that describes it. However, as noted above, $\mathbb{R}^n$ cannot be described by a single halfspace. Additionally, whenever another halfspace is added to your collection of halfspaces that describe $\mathbb{R}^n$, the number of possible points described cannot increase (that is, at best you retain the same number of points in the intersection). Since a single halfspace does not describe $\mathbb{R}^n$ and the addition of other halfspaces cannot add points to the description of $\mathbb{R}^n$, $\mathbb{R}^n$ cannot be described by the intersection of a finite number of halfspaces. Thus, $\mathbb{R}^n$ is not a polyhedron.