It is quite easy to give an example in real dimension $4$.

In fact, it was shown by D. Mumford in the paper

**The topology of normal singularities of an algebraic surface and a criterion for simplicity**, *Inst. Hautes Etudes Sci. Publ. Math.* (1961), no. 9, 5 -
22

that a normal algebraic surface over $\mathbb{C}$, whose underlying topological space (in the usual topology) is a topological manifold, must be nonsingular.

Now take the orbifold $X:=\mathbb{C}^2/G$, where $G$ is the group of order $2$ whose generator $\xi$ acts as $\xi \cdot (x, \, y) = (-x, \, -y)$.

The $G$-invariant subalgebra is generated by $A:=x^2, \, B=xy, \, C:=y^2$, so the underlying algebraic surface $X$ is isomorphic to the affine quadric cone $\{B^2-AC=0 \} \subset \mathbb{C}^3$, which has a $A_1$-singularity at its vertex. Such a singularity is normal, because it is a codimension two hypersurface singularity.

Then $X$ cannot be a topological manifold by Mumford's result mentioned above.

closedorbifold can be a manifoldwith boundary. For instance, the closed interval $[0,1]$ with both endpoints labelled $\mathbb{Z}/2$ is aclosed1-dimensional orbifold, since it's double covered by the circle. – HJRW Mar 8 at 11:20