For a radially symmetric wavefunction is $Psi$ allowed to blow up at $r=0$ provided that $|Psi|^2r^2$ doesn't?

Yes. Provided that the resulting wavefunction is normalizable. The point is that that, in polar coordinates the radial part of the Laplacian is a singular point of the equation at $r=0$. Depending on the form of the potential, such singular points can be in Weyl's limit point or limit circle case. In the latter case there is a one-parameter family of boundary conditions that can be imposed at $r=0$ for which the Schroedinger operator remains self adjoint. For three dimensions, and for the angular momentum $l>0$ case we are in the limit point case and the wavefunction has to remain finite. For the $l=0$ case we are in the limit circle case and the wavefunction is allowed to blow up provided we impose boundary conditions of the form $$ psi(r)sim Aleft(1-frac{alpha_s}{r}right),quad rto 0, $$ where $alpha_s$ is the scattering length. Physically, these conditions arise when there is something right at the origin which is far too small to be resolved by the wavelengths we are interested in. This happens in the case of cold atomic gases where $alpha_s$ is parametrizes the manner in which one atom perceives another.

1. From the point of view of the Hilbert space $H=L^2(mathbb{R}^3)$ itself the wave function is allowed to blow up not just in the origin $r=0$, but also in other points, as long as it is square integrable.

2. Now if the wavefunction should satisfy the TISE there are further restrictions.

3. If the TISE is spherically symmetric it is often useful to use spherical coordinates, as OP does. Additional conditions at $r=0$ is discussed in e.g. this & this Phys.SE posts.