The 0th Landau level also has a degeneracy of four as do other LLs. The significance of 0th LL is that it is shared by conduction and valence band with equal weight. That is to say: two of 0th LLs are electron like and two of them are hole-like (because the degeneracy is four, I imagine there are four "seperated" non-degenerate LLs at the same energy for pedagogical simplicity).
When an electron-like Landau-level is fully occupied by electrons, it contributes a unit conductance; when a hole-like Landau-level is fully occupied by holes, it contributes a negative unit conductance. (Assuming that the landau level has no degeneracy)
When an electron-like Landau-level is empty, it contributes no conductance (of course!); when a hole-like Landau-level is filled with electrons, it contributes no conductance.
Now let's see how the anomalous quantum hall effect arises:
When the 0th LL is fully occupied, the two electron-like LLs are filled with electron and two hole-like LLs are also filled with electron. Thus the conductance of the system is 2 unit conductance.
When the 0th LL is empty, the two electron-like LLs are empty and two hole-like LLs are filled with holes. Thus contribute to the conductance of -2 unit conductance.
Similarly, when both 0th and 1th LLs are also occupied with electrons, the total conductance is 4+2.
The origin of the anomalous Hall conductance
The anomalous Hall conductance arises because of the particle hole symmetry and the linear dispersion of graphene. For the 0th LL, because its energy is just 0, so it must be shared by the valence and conduction band.
However, for the parabolic dispersion, even if the original band have particle-hole symmetry. When forming LLs, the energy of lowest LLs formed by conduction band is 1/2, While the energy of lowest LL formed by valence band is -1/2. They do not mix.
Roughly speaking, the conductivity is inversely proportional to the scattering rate $1/\tau$ (in the Drude model, the longitudinal conductivity is $\sigma_0=\frac{n e^2 \tau}{m}$). From Fermi's golden rule, $1/\tau$ is proportional to the number of density of states a quasiparticle can scatter into, which is proportional to the density of states near the chemical potential. In the presence of a magnetic field, the energy levels are quantized into equally spaced Landau levels. Therefore the density of states at the chemical potential(or Fermi energy) $N(\mu)$ oscillates in $1/B$ periodically, where $B$ is the magnetic field, as the chemical potential passes each Landau level by varying the magnetic field. Therefore $1/\tau$ oscillates as well and the oscillation of $\sigma$ follows. That gives the SdH oscillation.
Note that this is different from the van-Alphen de-Haas oscillation, where thermodyanimc quantity such as magnetization is directly measured and the magnetization quantity is directly proportional to the DoS at the chemical potential. In the current case, the oscillation of DoS comes in indirectly via the scattering rate.
More detail discussions can be found in D. Schoenberg, Magnetic oscillations in metals.
Best Answer
In the infinite system limit, each Landau level is infinitely degenerate since the degeneracy scales with area.
In reality, no system is infinite. Tong is simply taking a finite system of size $L_x\times L_y$ as to quantize the momenta. This is the whole system under consideration, not only a small cut-out. All states are counted.
The result is that the degeneracy scales with $L_xL_y$. You may now choose the system size to be anything you want.