$2+1$-dimensional Einstein gravity is topological and only non-trivial globally
Let's start with a simpler example: $1+1$ dimensional gravity. This is actually a pretty important example to understand because all of string theory takes place in this framework!Within $1+1$ dimensional gravity, just as in $2+1$ dimensions, there are no local interactions. Thus, it seems like a pretty good starting place to talk about this kind of stuff.
If we denote by $h$ our metric tensor and by $R$ the scalar curvature of our manifold $\mathcal{M}$, then we can write
$$S=\frac{1}{2}\int_{\mathcal{M}}\mathrm{d}^2\sigma\,\sqrt{h}\,R,$$
where here $h$ denotes the scalar $\det{h^{\alpha\beta}}$ and $\sigma$ is our choice of coordinates on open patches of $\mathcal{M}$. There is a beautiful theorem, known as the Gauss-Bonnet theorem, which states that this integral can be solved without the knowledge of the coordinates or the metric. It states that
$$\int_{\mathcal{M}}\mathrm{d}^2\sigma\,\sqrt{h}\,R=4\pi\chi(\mathcal{M}),$$
where $\chi(\mathcal{M})$ is the so-called Euler characteristic of the manifold. This is a purely topological expression. Intuitively, it counts the number of "holes" in a continuous manifold (namely, $\chi(\mathcal{M})=2-2g$, where $g$ is the number of holes in the manifold -- for instance, a torus has one hole, and so has $\chi=0$, while a sphere has no holes and has $\chi=2$). Thus, since the action itself is independent of local features of the manifold but is dependent of the global (read: topological) features, we say that $1+1$ dimensional gravity is locally trivial but globally nontrivial.
The confusion may be caused by your (understandable) thinking that gravity is all about forces between objects. But in a modern sense, gravity is identical to geometry. This includes topological statements as well as local ones! We are used to talking about gravity as a local thing because the standard treatment of GR is a local one. But gravity determines the topology of our manifold as well as the local geometry!
This is all good and well, but why do we care? Well, in $1+1$ dimensional quantum gravity (read: string theory), the topology plays an important role! This is because the path integral treatment, we have (in a Euclidean signature),
$$\langle\mathcal{O}(\phi)\rangle=\sum_{\text{g=0}}^{\infty}\int\mathcal{D}\phi\,\mathcal{D}h\,\mathcal{O}(\phi)\,e^{-S[\phi,h]},$$
where $\phi$ is a placeholder for all of the fields propagating on the manifold, $\mathcal{O}$ is some operator function of our fields, and
$$S[\phi,h]=\int\mathrm{d}^2\sigma\sqrt{h}\left(-\frac{\lambda}{8\pi}R+\mathcal{L}(\phi,h)\right),$$
where $\lambda$ is an effective coupling constant and $\mathcal{L}(\phi,h)$ is the Lagrangian that defines the gravitational interaction of the fields $\phi$. The sum over $g$ indicates that we are summing over all possible topologies of the theory by counting over the genus. Using the Gauss-Bonnet theorem, we can immediately write down
$$\langle\mathcal{O}(\phi)\rangle=\sum_{g=0}^{\infty}e^{\lambda(1-g)}\int\mathcal{D}\phi\,\mathcal{O}(\phi)\,e^{-S_g[\phi]}.$$
Assuming $\lambda$ is positive, this tells us that we can essentially treat expectation values as being perturbative expansions in possible topologies of our manifold!
While I didn't really talk about $2+1$ dimensional gravity, the basic takeaway should be this:
While $2+1$ gravity doesn't have local degrees of freedom, the topology is still a geometric property of the spacetime manifold $\mathcal{M}$, and is thus still important and nontrivial.
Global nontrivialities can play an important role in the quantum version of the gravitational theory you're working in.
Finally, to answer your question in one sentence (as if I haven't babbled on enough): $2+1$ dimensional Einstein gravity is called "topological" because its only possible (vacuum) nontrivialities are all topological in nature.
[Note: throughout I assumed a vacuum or "pure" gravitational theory. Einstien gravity in $2+1$ dimensions with point masses is actually very interesting and has the same geometry as a cone. This can lead to some very interesting and unintuitive results, the most famous of which is the Gott time machine (worth a read).]
I hope this helped!
Bob Knighton's answer is very detailed. But I want to add a few remarks. You can try to prove this identity
$$ R_{\mu\nu\rho\sigma} = \frac{R}{D(D-1)} (g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}). $$
Then, it's straightforward to show that for 1+1 spacetime, the Einstein tensor $R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}$ is identically zero. That is, the vacuum Einstein field equation is trivially satisfied, no matter how much matter content you put. This means, there is no coupling between gravity and matter.
Similar consequences also happen on the 2+1 case. In fact, people had believed that 2+1 gravity was just as trivial as 1+1 gravity. However, in around 1990, it was discovered that in presence of a negative cosmological constant, 2+1 gravity is a rich subject (there are BTZ black hole solutions, etc).