I will approach this question theoretically, although I feel the intuition follows nicely. If we talk about Kerr black holes - rotating black holes described by their mass and angular momentum, with no additional parameters such as charge etc. - then you can show that the radius of the event horizon is given by
$\boxed{r=M + \sqrt{M^2-a^2}}$
where $a=\frac{J}{M}$.
(This value of $r$ is found by finding where the Kerr metric blows up; hence event horizon. In fact, finding where the metric blows up involves solving a quadratic equation, so we get two values of $r$ and in Kerr black holes we therefore have two event horizons; unlike in Schwarzschild black holes.)
Regarding your first point about maximum angular momentum, if we set $G=1$ and $c=1$, the maximum angular momentum you stated is given by $a=M$ and if we plug this into our equation for $r$ above we see that we have
$r=M$.
We know that the radius of the event horizon in a Schwarzschild black hole (no rotation) is $r=2M$. So therefore we can see that at maximum angular momentum, the radius of the event horizon is half of what it would be if the black hole weren't spinning.
To this end, we can also see that at zero angular momentum, $a=0$, we have
$r=2M$
which is what we want as at zero angular momentum we of course should have the Schwarzschild radius.
Using the boxed equation for $r$ at the top, it's easy to test out different values of $a$ to see what happens to the event horizon. For example, this equation alone is sufficient to show that for $a>M$ we don't have an event horizon, in which case we have what is a called "Fast Kerr" which is just a singularity with no event horizon.
Best Answer
The Schwarzschild metric is given by,
$$\mathrm{d}s^2 = \left( 1-\frac{2GM}{r}\right)\mathrm{d}t^2 -\left( 1-\frac{2GM}{r}\right)^{-1}\mathrm{d}r^2 -\underbrace{r^2\mathrm{d}\theta^2 -r^2 \sin^2 \theta \mathrm{d}\phi^2}_{r^2\mathrm{d}\Omega^2}$$
in spherical coordinates. Notice the metric tensor is singular at the origin, $r=0$, and at the Schwarzschild radius $r=2GM$. The former is a true physical curvature singularity. However, as one can verify by computing various curvature scalars, the singularity $r=2GM$ is unphysical, and can be removed by a coordinate transformation. In particular, the metric in Kruskal coordinates is given by,
$$\mathrm{d}s^2 = \frac{32G^3M^3}{r}e^{-r/2GM} \left( \mathrm{d}U^2 -\mathrm{d}V^2\right) + r^2 \mathrm{d}\Omega^2$$
The event horizon $r=2GM$ (in natural units) corresponds to $V=\pm U$, and indeed the metric is not singular at the point, reflecting the fact the singularity arose simply because we chose an inappropriate coordinate system. Regarding the $r=0$ singularity, we cannot define a notion of length or associated any length to the singularity; it is a single point on the manifold.
At the Planck scale, we cannot resort solely to general relativity, and quantum gravity effects become important. From a quantum field theory perspective, this is typical. Short distances correspond to a high energy scale where our theory does not provide a description. As Prof. Tong states,
Nevertheless, gravity is different from simply an effective field theory, such as Fermi's theory of the weak interaction. We may still make predictions regarding gravity for high energies; e.g. if we collide at energies above the Planck mass, we know we form a black hole.
A final remark as other answers mentioned naked singularities. When quantum gravity effects become important, after we pass the event horizon, we cannot communicate them to others behind the horizon; this is an example of cosmic censorship. However, violations have been hypothesized, c.f. the Gregory-Laflamme instability of $p$-branes and black strings.