How does intuition fail for higher dimensions? From this answer:

Now, Algebraic Geometry is one of the oldest, deepest, broadest and
  most active subjects in Mathematics with connections to almost all
  other branches in either a very direct or subtle way. The main
  motivation started with Pierre de Fermat and René Descartes who
  realized that to study geometry one could work with algebraic
  equations instead of drawings and pictures (which is now fundamental
  to work with higher dimensional objects, since intuition fails there).

What are these failings? Can you give me some examples?
 A: Spheres in a hypercube
Here is one example.

Take a square of edge length $4$. Place $2^2=4$ circles of radius $1$ into that square, in the obvious arrangement. Then place a circle at the center of all these circles, and make that fifth circle as large as possible. To compute its radius, look at the distance between the center of the square and the center of one of the large circles. That difference vector is $1$ length step for each of your two dimensions, so the distance is $\sqrt2$. As the big circle has radius $1$, your little circle has radius $\sqrt2-1$.
Now go to dimension $n$. You have a hypercube of edge length $4$, into which you place $2^n$ spheres of radius $1$ and one central sphere of radius $\sqrt n-1$. For $n=4$, the central sphere will already be as big as the outer spheres, and for $n>4$ it will be larger. At $n=9$, the radius of the central sphere will become $2$ so the central sphere will touch the containing hypercube. For $n>9$ the central sphere will no longer fit into the hypercube, even though it still touches the insides of all the radius-1-spheres contained in the hypercube.
This seems pretty counter-intuitive to me, even though the algebra is simple enough.
Finite sphere packings
A remotely related phenomenon: take a finite number $m$ of unit spheres, and arrange them in such a way that the volume of the convex hull becomes minimal. You may imagine that for a few spheres, placing them in a straight line will be best, but the more spheres you have the more compact a clustering would appear in comparison. For 3D, more compact packings exist for e.g. $m=56$ spheres.
For $n\geq42$ dimensions (note that number!) it has been proven that the “sausage conjecture” holds: for those dimensions, the straight line sausage arrangement will always be optimal no matter the number of spheres. It has been conjectured but not proven that the sausage is always optimal for $n\geq5$ dimensions. The German Wikipedia article on this subject has some nice illustrations, even if you don't understand the language.
A: No algebraic geometry in it but an example worth knowing concerns high-dimensional spheres. Roughly speaking the result is that, when the dimension is high enough, all the mass of the sphere is concentrated on anyone of its equators. A bit as if every $\varepsilon$-neighborhood of the Equator was enough to encompass nearly all the surface of the Earth. As every $\varepsilon$-neighborhood of each pair of opposite meridians...
Rigorously speaking, one introduces the uniform probability measure $u_n$ on the $n$-dimensional sphere $S^n=\{x=(x_k)_{1\leqslant k\leqslant n+1}\in\mathbb R^{n+1}\mid x_1^2+\cdots+x_{n+1}^2=1\}$ and the $\varepsilon$-neighborhood $N^n_\varepsilon$ of the great circle $x_{1}=0$, defined by $N^n_\varepsilon=\{x\in S^n\mid |x_{1}|\leqslant\varepsilon\}$. Then, for every $\varepsilon\gt0$, $u_n(S^n\setminus N^n_\varepsilon)\to0$ when $n\to\infty$.
Note that, by symmetry, the same applies to $\bar N^n_\varepsilon=\{x\in S^n\mid |x_{42}|\leqslant\varepsilon\}$. 
A: Intuition already breaks down at the fact that two planes can meet in a single point. The moral is that visualizing intersections in higher dimensions is hard.
Another example: the volume of the $n$-sphere. It starts at 2, then increases until dimension 4, then it decreases again, and in fact, goes to zero. (there is in fact a very long discussion of this on mathoverflow)

