Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.
The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.
- The zero-tree is the one-point space.
- A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.
I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.
Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.
Edit: As Gera points out, the annulus is an easy counterexample to my claimed “Lemma”. I now do not see how to strengthen the OP hypotheses without making them very combinatorial… If you are willing to do that, then Ian’s answer seems like a very natural direction to go in.