Are most matrices diagonalizable?

Yes. Here is a proof over $\mathbb{C} $.

• Matrices with repeated eigenvalues are cut out as the zero locus of the discriminant of the characteristic polynomial, thus are algebraic sets.
• Some matrices have unique eigenvalues, so this algebraic set is proper.
• Proper closed algebraic sets have measure $0.$ (intuitively, a proper closed algebraic set is a locally finite union of embedded submanifolds of lower dimension)
• (over $\mathbb{C} $) The set of matrices that aren't diagonalizable is contained in this set, so it also has measure $0$. (not over $\mathbb{R}$, see this comment https://math.stackexchange.com/a/207785/565)

Let $A$ be a real matrix with a non-real eigenvalue. It's rather easy to see that if you perturb $A$ a little bit $A$ still will have a non-real eigenvalue. For instance if $A$ is a rotation matrix (as in Georges answer), applying a perturbed version of $A$ will still come close to rotating the vectors by a fixed angle so this perturbed version can't have any real eigenvalues.