I don't know if this is the kind of answer you expect, but:
In the hyperbolic space of dimension $n+1$ one naturally gets all $n$-dimensional constant curvature geometries.
- spheres (points at distance $\le r$ from a given point) inherit their spherical $n$-dimensional geometry. I'm not sure what the curvature is as a function of $r$, but it tends to $0$ resp. $\infty$ when the radius tends to $\infty$ resp. $0$.
- horospheres inherit the Euclidean $n$-dimensional metric. Recall that a horofunction is a limit $h$ of functions $x\mapsto d(x,x_n)-d(x_0,x_n)$ for some sequence $x_n$ tending to infinity, and a horosphere is $\{x:h(x)=0\}$ for some horofunction $h$.
- for two distinct points $x_1,x_2$, the set of $x$ such that $d(x,x_1)=d(x,x_2)$ inherits the $n$-dimensional hyperbolic geometry.
Edit: One can even fully interpolate: consider the half-space model $H$ in $\mathbf{R}^{n+1}$. For a round sphere or hyperplane $S$ in $\mathbf{R}^{n+1}$ with $S\cap H\neq\emptyset$, the intersection $S\cap H$ achieves all these cases: either $S\subset H$, hence is a sphere in $H$ as well (not with the same center!), with all possible positive curvature. If $S$ is tangent to $\partial H$ (or $S$ is parallel to $H$), then it is a horosphere, i.e. with zero curvature. If $S$ is not contained in $\bar{H}$, then one gets something with constant negative curvature (which is totally geodesic when $S$ meets $\partial H$ in a perpendicular way). It seems however that the negative curvature cannot be arbitrarily close to $-\infty$, by a compactness argument. Possibly it is $\ge -1$ with equality exactly in the geodesic case?