If we detect N photons from a coherent state of light for a measurement,

Short answer: A good order of magnitude rule of thumb for the maximum possible bandwidth of an optical fibre channel is about 1 petabit per second per optical mode. So a "single" mode fibre (fibre with one bound eigenfield) actually has in theory two such channels, one for each polarisation state of the bound eigenfield.

I'll just concentrate on the theoretical capacity of a single, long-haul fibre; see roadrunner66's answer for discussion of the branching in an optical network. The fundamental limits always get down to a question of signal to noise in the measurement (i.e. demodulation by the receiver circuit). The one, fundamentally anavoidable, noise source on a fibre link is quantum shot noise, so I'll concentrate on that. Therefore, what follows will apply to a short fibre: other noise sources (such as Raman, amplified spontaneous emission from in-line optical amplifiers, Rayleigh scattering, Brillouin scattering) tend to become significant roughly in proportion to the fibre length and some power (exponent) of the power borne by the fibre.

If we detect N photons from a coherent state of light for a measurement, then the undertainty in that measurement is N−−√$\sqrt{N}$ in an otherwise noise free environment. (see footnote on squeezed states). So suppose first that:

1. The bandwidth of our fibre is B$B$ hertz (typically 50THz maximum, I'll discuss what limits B$B$ below)
2. The power borne by the fibre is P$P$ watts (typically 100W maximum; again, I'll discuss limits later)
3. The fibre is single mode (this in theory allows one to overcome the dispersion limits discussed in roadrunner66's answer, at the expense of putting a harder upper limit on P$P$)
4. The fibre's centre frequency is ν0${\nu }_0$ (typically 193THz, corresponding to 1550nm freespace wavelength)
5. In what follows, I shall take the word "symbol" to mean a theoretically infinite precision real number but whose precision is limited by noise (the purpose of this answer being to discuss the latter!).

So, let's begin by exploring the best way to use our power. If we devote it to M$M$ symbols per second, each of our measurements comprise the detection of N=PMhν0$N=\frac{P}{Mh{\nu }_0}$ photons, thus our signal to noise ratio is SNR=NN√=PMhν0−−−−−√$SNR=\frac{N}{\sqrt{N}}=\sqrt{\frac{P}{Mh{\nu }_0}}$. By the Shannon-Hartley form of the Noisy channel coding theorem (see also here), we can therefore code our channel to get log2(1+PMhν0−−−−−√)${\log }_2(1+\sqrt{\frac{P}{Mh{\nu }_0}})$ bits of information per symbol, i.e. Mlog2(1+PMhν0−−−−−√)$M{\log }_2(1+\sqrt{\frac{P}{Mh{\nu }_0}})$ bits per second through our optical fibre. This is a monotonically rising function of M$M$, so a limit on P$P$ by itself does not limit the capacity.

However, by a converse of the Nyquist-Shannon sampling theorem we can send a maximum of B$B$ symbols down the channel per second. This then is our greatest possible symbol rate. Hence, our overall expression for the fibre capacity in bits is:

C=Blog2(1+PBhν0−−−−√)$\mathcal{C}=B{\log }_2(1+\sqrt{\frac{P}{Bh{\nu }_0}})$ bits per second

If we plug in our B=50THz$B=50\mathrm{T}\mathrm{H}\mathrm{z}$ and P=100W$P=100W$, we get the stupendous capacity of 1.2 petabits per second for each single mode optical fibre core.

By looking at the basic shape of the Shannon-Hartlee expression log2(1+SNR)${\log }_2(1+SNR)$ bits per symbol, one can see that improvements on the signal to noise beyond any "good" signal to noise ratio will only lead to marginal increases in capacity. By far the strongest limit is the Shannon sampling theorem converse. So the use of squeezed light will not change the order of magnitude of this result.

Now for some material physics begetting the limits discussed above. Our optical power is limited by two things:

1. The heat dissipation rate of the fibre is the main one. At some point, losses in the fibre sum up to more power than it can dump to its surroundings and it fries itself. What happens in practice is that power tends to dissipate around inclusions and other imperfections and then it melts at that point, more power dissipates at the molten blob and the destruction propagates slowly away from the failure: the fibre "melts" itself like a lit fuse.
2. Nonlinear processes like stimulated Brillouin and stimulated Raman scattering will quickly outweigh the quantum noise. These vary like P2$P^2$.
3. cable technician certification

Bandwidth is limited by the losses in the medium. The "window" between about 1300nm and 1600nm freespace wavelength is chosen for optical communications for its low loss. Depending on the length of fibre, as you try to increase the bandwidth, any optical power outside the band of low loss simply won't get to the other end. This is what gives rise to the B=50THz$B=50THz$ figure I cite above. This is not a hard figure: it's a rough estimate and its precise value depends on the length of fibre. I have shown a calculation where I assume the fibre transmits a certain bandwidth perfectly and sending "switches off" altogether outside this bandwidth. A fuller calculation would account for the spectral shape of the losses and that the imperfectly sent frequency components can also bear information. This would show that the effective equivalent bandwidth B$B$ in my formula would be roughly inversely proportional to fibre length.