If you roll a sheet of paper so left and right touch, then bend the cylinder so its ends also touch, you can see the surface of a 2D rectangle maps onto the surface of a 3D torus, a doughnut. I was wondering if there is a name for the topological object which results when you do a similar mapping of a 3D cube ? I would guess its in 4D ?
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This is known as the three-torus.
When you roll a two-dimensional sheet of paper into a two-torus, you are taking the Cartesian product of two intervals, denoted $I \times I$, and wrapping each interval separately into circles, which we denote $S^1$, the one-sphere. This is why you see the two-torus notated as $S^1 \times S^1$. Similarly, the procedure of "folding" a cube converts $I \times I \times I$ into $S^1 \times S^1 \times S^1$. Obviously this procedure generalizes for arbitrary $n$.
As NKS says, this surface is the three-torus. To answer the question of whether it's 'in 4D' or not, though, requires a bit of definition as to what that phrase means. The torus itself, for instance, isn't inherently a 3-dimensional object; as the construction suggests, the torus's surface is essentially 2-dimensional in that around any given point on the surface you can parametrize it with just two coordinates. When we say it's 'three-dimensional' what's really meant (generally) is that it's embeddable into 3-dimensional space; that is, that there's a continuous map from points on the surface of the torus to points in 3-dimensional space such that no two points on the torus map to the same point in 3-space.
Now, not all 2-dimensional surfaces can be embedded in 3-dimensional space; the Klein Bottle is the most common example of one that can't. On the other hand, it turns out that every 2-dimensional surface (the formal term is a 2-dimensional manifold) can be embedded in 4-dimensional space. In general, $n$-dimensional manifolds can be embedded in $(2n)$-dimensional space, but there's no guarantee that they can be embedded in a space with fewer dimensions; so the worst-case scenario for the '3D torus' would be that it embedded into 6-dimensional space. Fortunately, it can be easily shown that this doesn't happen for the torus; your 3D torus can be embedded in 4D space. To see this, consider the process of going from a circle (the '1D torus') to a standard 2D torus - as NKS notes, this is essentially the process of 'wrapping' one circle around another, by (for instance) taking a circle in the $x$-$y$ plane with its center on the $x$ axis (and not containing the origin) and rotating it about the $y$ axis so that the circle's center sweeps out another circle on the $x-z$ plane. Similarly, you can take a torus with its center on the $x$ axis (and removed from the origin) and sweep it about the '$yz$ plane' so that the center of the torus sweeps out a circle on the $x-w$ plane; with a little work you can convince yourself that this is an embedding (because only two points in the 'sweep interval' have the same $w$ value, and their copies of the torus don't intersect because we started our torus offset from the origin). Likewise, by continuing to sweep out more circles we can embed the 4D torus in 5D space, the 5D torus in 6D space, etc.