Let $PD_{n}$ be the cone of positive definite $n \times n$ real matrices and let $B$ be the unit sphere in $n \times n$ dimensions. What is the volume of $PD_{n} \cap B$?
EDIT: Let's assume that $B$ is the unit sphere w.r.t the operator norm: $||A||=\sup_{||x||=1}{||Ax||}$.
Using the operator norm, as you have defined it, the fraction of the unit ball in real symmetric $n$-by-$n$ matrices that consists of positive definite matrices is $2^{-n(n+1)/2}$. Thus, this fraction shrinks very quickly as $n$ increases.
Added comment 1: Actually, I just realized that you don't need to do the calculation below; the ratio is obvious anyway: Consider the map $\Phi:B_n\to \overline{PD_n\cap B_n}$ defined by $\Phi(A) = \tfrac12(A+I_n)$. This affine invertible map identifies $B_n$ with $\overline{PD_n\cap B_n}$, which has the same volume as $PD_n\cap B_n$. Obviously, it multiplies volumes by a factor of $2^{-n(n+1)/2}$, since $\tfrac12n(n{+}1)$ is the dimension of the ambient space of symmetric $n$-by-$n$ matrices. QED.
Added comment 2: I note that the OP never did say that he wanted to concentrate on symmetric positive definite matrices, which seems bizarre to me, but maybe he really does mean this. I looked online and found that some sources say that a real $n$-by-$n$ matrix $A$ is positive definite if $x\cdot Ax \ge 0$ for all $x\in\mathbb{R}^n$, with equality if and only if $x = 0$. (Of course, this is a condition only on the symmetric part of $A$.) If one does adopt this convention and considers the problem as being posed on the set of all $n$-by-$n$ real matrices with the operator norm, then the problem becomes much harder, and the above scaling trick doesn't work. I don't know the answer in this case for general $n$, but, for $n=2$, there is a fortuitous simplification: If $$ A = \begin{pmatrix}a^1_1&a^1_2\\ a^2_1&a^2_2\end{pmatrix} = \begin{pmatrix}r+a&b+s\\ b-s& r-a\end{pmatrix}, $$ then, for the operator norm as defined by the OP, one has a nice formula $$ \| A\| = \sqrt{r^2+s^2} + \sqrt{a^2+b^2} $$ and $A$ is positive definite if and only if $r>\sqrt{a^2+b^2}$. (For higher values of $n$, these are not nice at all.)
Now, it's just a calculus problem. The volume of the unit ball turns out to be $V = 2\pi^2/3$, and the volume of the positive definite matrices turns out to be $$ V_+ = 4\pi(53-15\pi)/45 $$ with the resulting ratio being $$ V_+/V \approx 0.2493898621\ . $$ Thus, nearly one-quarter of the unit ball in this case consists of positive definite matrices.
Here is how the calculation goes: First, we need to parametrize this unit ball in some way. The easiest way to do this is to write $$ A = g R g^{-1} $$ where $g$ is an element of the orthogonal group $\mathrm{O}(n)$ and $R=\mathrm{diag}(r_1,\ldots,r_n)$ where the the $r_i$ belong either to the domain $$ B = \{(r_1,\ldots,r_n)\ |\ 1\ge r_1\ge r_2\ge\cdots\ge r_n\ge-1\} $$ (in which case $A$ belongs to the unit ball in the operator norm) or $$ B_+ = \{(r_1,\ldots,r_n)\ |\ 1\ge r_1\ge r_2\ge\cdots\ge r_n\ge0\} $$ (in which case $A$ belongs to the part of the unit ball that consists of positive semidefinite symmetric matrices). (Actually, this parametrization is generically $2^n$-to-$1$ from $B\times\mathrm{O}(n)$ to the unit ball in the operator norm and $2^n$-to-$1$ from $B_+\times\mathrm{O}(n)$ to the postive semi-definite matrices within the unit ball, but that won't matter to the calculation of the ratio of volumes.)
Now, a straightforward calculation using this parametrization shows that the two volumes we want to compute are $\Omega_n$ (the volume of $\mathrm{O}(n)$) divided by $2^n$ times the integrals over $B$ and $B_+$ of the following $n$-form: $$ \phi = \left(\prod_{i<j}(r_i-r_j)\right)\, \mathrm{d}r_1\wedge\mathrm{d}r_2\wedge\cdots\wedge \mathrm{d}r_n $$ Fortunately, it is not necessary to compute this integral over either domain because, if you look at the mapping $f:B_+\to B$ defined by $$ f(r_1,r_2,\ldots,r_n) = (2r_1{-}1,2r_2{-}1,\ldots,2r_n{-}1), $$ you see immediately that $$ f^\ast\phi = 2^{n(n+1)/2}\,\phi $$ Thus, the integral of $\phi$ over $B_+$ divided by the integral of $\phi$ over $B$ is $ 2^{-n(n+1)/2}$, as claimed.
