So in 3-D the differential equation $$\ddot{\bf{r}} = -\frac{f(r)}{m}\bf{r}$$
is shown to be planar by noting $$\bf{r} \times \dot{\bf{r}}$$
is constant. But isn't the differential equation planar (lie in a 2-d subspace) in n-D? If so, how do you show that?
Yes, the motion still takes place in a plane (the plane spanned by $\mathbf{r}(0)$ and $\mathbf{\dot r}(0)$).
One way you can see it is by expanding $\mathbf{r}(t)=\mathbf{r}(0)+\mathbf{\dot r}(0)t+...$ in a series and using the differential equation to see that $\mathbf{r}^{(k)}(0)\in $ span$\{ \mathbf{r}(0), \mathbf{\dot r}(0)\}$.
The other way is the angular momentum way. To generalize cross product you use wedge product. The wedge of two vectors $x, y$ is the skew symmetric map: $x\wedge y:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ given by $(z,w)\mapsto (z\cdot x)(w\cdot y)-(w\cdot x)(z\cdot y)$.
Then given a plane $\sigma=$ span$\{x,y\}\subset \mathbb{R}^n$ the complement $\sigma^\perp = \{ z\in \mathbb{R}^n : x\wedge y (z, \cdot)\equiv 0\}$.
Now you differentiate $\mathbf{r}\wedge \mathbf{\dot r}$ along a solution:
$\frac{d}{dt}\mathbf{r}\wedge \mathbf{\dot r}=\mathbf{\dot r}\wedge \mathbf{\dot r}+\mathbf{r}\wedge \mathbf{\ddot r}=-\frac{f(r)}{m}\mathbf{r}\wedge \mathbf{r}=0$
so it is the same map for all $t$ so that the complement to the plane of motion is constant i.e. the plane of motion is constant.