Are most matrices invertible?

There are at least three ways of saying that a matrix over the real numbers is generically invertible:

The topological one: the set of invertible matrices is a dense open set in the set of all matrices.

The probabilistic one: with the Lebesgue measure on the set of matrices, the non-invertible matrices are of measure zero.

The algebraic one: The set of invertible matrices is open (and non-empty) for the Zariski topology; explicitly, this means that there is a polynomial defined on the coefficients of the matrices, such that the set of invertible matrices is exactly the set where this polynomial is not zero. Of course, here the polynomial is the determinant function.

Remark that your question makes sense for matrices with coefficients in an arbitrary infinite field (for finite fields, we are looking at finite sets...) and that we can still say that a matrix is generically invertible in the algebraic sense.

Yes. The non-invertible matrices form a set of measure zero in the space of all $n\times n$ matrices, when considered as a subset of $\mathbb R^{n^2}$.

To expand on Potato's answer, the space of all $n \times n$ matrices is manifold isomorphic to $\mathbb{R}^{n^2}$. The determinant gives an polynomial function from this space to $\mathbb{R}$, and the zero set of a polynomial function has measure $0$.

(Earlier this asserted that the set of singular matrices is a manifold, which is not true.)