What's new in higher dimensions?
In high dimensions, almost all of the volume of a ball sits at its surface. More exactly, if $V_d(r)$ is the volume of the $d$-dimensional ball with radius $r$, then for any $\epsilon>0$, no matter how small, you have $$\lim_{d\to\infty} \frac{V_d(1-\epsilon)}{V_d(1)} = 0$$ Algebraically that's obvious, but geometrically I consider it highly surprising.
Edit:
Another surprising fact: In 4D and above, you can have a flat torus, that is, a torus without any intrinsic curvature (like a cylinder in 3D). Even more: You can draw such a torus (not an image of it, the flat torus itself) on the surface of a hyperball (that is, a hypersphere). Indeed, the three-dimensional hypersphere (surface of the four-dimensional hyperball) can be almost completely partitioned into such tori, with two circles remaining in two completely orthogonal planes (thanks to anon in the comments for reminding me of those two leftover circles). Note that the circles could be considered degenerate tori, as the flat tori continuously transform into them (in much the same way as the circles of latitude on the 2-sphere transform into a point at the poles).
A number of problems in discrete geometry (typically, involving arrangements of points or other objects in $\mathbb R^d$) change behavior as the number of dimensions grows past what we have intuition for.
My favorite example is the "sausage catastrophe", because of the name. The problem here is: take $n$ unit balls in $\mathbb R^d$. How can we pack them together most compactly, minimizing the volume of the convex hull of their union? (To visualize this in $\mathbb R^3$, imagine that you're wrapping the $n$ balls in plastic wrap, creating a single object, and you want the object to be as small as possible.)
There are two competing strategies here:
- Start with a dense sphere packing in $\mathbb R^d$, and pick out some roughly-circular piece of it.
- Arrange all the spheres in a line, so that the convex hull of their union forms the shape of a sausage.
Which strategy is best? It depends on $d$, in kind of weird ways. For $d=2$, the first strategy (using the hexagonal circle packing, and taking a large hexagonal piece of it) is best for almost any number of circles. For $d=3$, the sausage strategy is the best known configuration for $n \le 56$ (though this is not proven) and the first strategy takes over for larger $n$ than that: the point where this switch happens is called the "sausage catastrophe".
For $d=4$, the same behavior as in $d=3$ occurs, except we're even less certain when. We've managed to show that the sausage catastrophe occurs for some $n < 375,769$. On the other hand, we're not even sure if the sausage is optimal for $n=5$.
Finally, we know that there is some sufficiently large $d$ such that the sausage strategy is always the best strategy in $\mathbb R^d$, no matter how many balls there are. We think that value is $d=5$, but the best we've shown is that the sausage is always optimal for $d\ge 42$. There are many open questions about sausages.
If you're thinking about the more general problem of packing spheres in $\mathbb R^d$ as densely as possible, the exciting stuff also happens in dimensions we can't visualize. A recent result says that the $E_8$ lattice and the Leech lattice are the densest packing in $\mathbb R^8$ and $\mathbb R^{24}$ respectively, and these are much better than the best thing we know how to do in "adjacent" dimensions. In a sense, this is saying that there are $8$-dimensional and $24$-dimensional objects with no analog in $\mathbb R^d$ for arbitrary $d$: a perfect example of something that happens in many dimensions that can't be intuitively described by comparing it to ordinary $3$-dimensional space.
Results like the Hales–Jewett theorem are another source of "new behavior" in sufficiently high-dimensional space. The Hales–Jewett theorem says, roughly speaking, that for any $n$ there is a dimension $d$ such that $n$-in-a-row tic-tac-toe on an $n \times n \times \dots \times n$ board cannot be played to a draw. (For $n=3$, that dimension is $d=3$; for $n=4$, it's somewhere between $d=7$ and $d = 10^{11}$.) However, you could complain that this result is purely combinatorial; you're not doing so much visualizing of $d$-dimensional objects here.