Yes, you can form a black hole from just photons. There is a fairly simple solution that captures exactly what you are describing, given on pg. 83 of Eric Poisson's GR lecture notes: http://www.physics.uoguelph.ca/poisson/research/agr.pdf.
Basically, the setup is that you have a thin sphere of energy density collapsing at the speed of light, which could be taken to be your spherical arrangement of photons. Inside, the metric is flat space,
$$ds^2 = -dt^2_-+dr^2 + r^2 d\Omega^2,$$
and outside the sphere it is the Schwarzschild solution
$$ds^2 = -fdt^2_++\frac1fdr^2+r^2d\Omega^2,$$
with $f=1-2M/r$, and $M$ constant. Here $t_-$ and $t_+$ don't necessarily coincide at the boundary of the sphere, but the spatial coordinates $(r,\theta, \phi)$ do coincide at the boundary. From the inside, the surface of the photon shell is located at $t_-=v_--r$, while from the outside the boundary is $t_+=v_+-r_*$, where $v_-, v_+$ are constants and $r_*=r+2M\ln(r/2M-1)$ is the tortoise coordinate.
The notes go on to calculate the surface energy density and pressure, and unsurprisingly find that the energy density is $\mu = M/4\pi r^2$ and the pressure is zero.
It's also worth noting that you don't really need a large number of photons for this to work: as long as you have them collapsing in a perfectly spherical, very thin shell, they will form a black hole once the shell becomes smaller than the Schwarzschild radius. Fewer (or less energetic) photons will simply make a smaller black hole.
Note that as a practical matter, you would be limited by how thin you can actually make the shell. In order for the thin shell approximation to work, you need the actual width of the shell (or say the photon wavepackets) to be much less than the Schwarzschild radius of the eventual black hole that will form. For laboratory lasers, the Schwarzschild radius will be way too tiny to ever make a black hole using this setup.
