Along the adiabatic paths, $$\Delta Q=0$$ so $$dU=-pdV$$ meaning the work done is given by $$W=-\Delta U$$.
This changes the values I proposed in my question.
In fact:
Along the 2 --> 3 transition $$W=U_{2}-U_{3}=a(T_{C}^{4}V_{2}-T_{H}^{4}V_{3})$$
Along the 4 --> 1 transition $$W=U_{4}-U_{1}=a(T_{H}^{4}V_{4}-T_{C}^{4}V_{1})$$
This correction yields $$\eta=\frac{a\left[T_{H}^{4}\left(V_{4}-V_{3}\right)+T_{c}^{4}\left(V_{2}-V_{1}\right)\right]+\dfrac{1}{3}aT_{H}^{4}(V_{4}-V_{3})+\dfrac{1}{3}a(T_{C}^{4}(V_{2}-V_{1}))}{\dfrac{4}{3}aT_{H}^{4}(V_{4}-V_{3})}$$
And since $T^{3}V=constant$ along adiabatic curves, we can use the substitutions: $$V_{1}=(\frac{T_H}{T_{C}})^{3}V_{4}$$ and $$V_{2}=(\frac{T_H}{T_{C}})^{3}V_{3}$$ to solve for our final answer, which is the expected Carnot efficiency: $$\eta=\frac{T_{H}-T_{C}}{T_{H}}$$
The Short Answer
How is the efficiency of a heat engine related to the entropy produced during the process?
The maximum efficiency for any heat engine operating between two temperature $T_H$ and $T_C$ is the Carnot efficiency, given by
$$e_C = 1 -\frac{T_C}{T_H}.$$
Such a heat engine produces no entropy, because we can show that the entropy lost by the hot reservoir is exactly equal to the entropy gain of the cold reservoir, and of course, the system's entropy on the net doesn't change because the system undergoes a cycle.
Any heat engine operating between the same two temperatures whose efficiency is less than $e_C$ necessarily increases the entropy of the universe; in particular, the total entropy of the reservoirs must increase. This increase in entropy of the reservoirs is called entropy generation.
Finally, the efficiency of the perfect engine is less than one, necessarily, because the entropy "flow" into the system from the hot reservoir must be at least exactly balanced by the entropy "flow" out of the system into the cold reservoir (because the net change in system entropy must be zero in the cycle), and this necessitates waste heat from the system into the cold reservoir. The fact that $e_C$ goes to one in the limit of small ratios $T_C/T_H$ is a consequence of the fact that $Q_C$ is small compared to $Q_H$. It is not a consequence of the fact that entropy generation is small in this case, because entropy generation is already zero for the Carnot cycle.
Explanation
Let's concentrate first on the interaction between the system and the hot reservoir. An amount $\delta Q_H$ of energy flows into the system from the hot reservoir, which means that the system's entropy changes by
$$\mathrm dS_\text{sys} = \frac{\delta Q_H}{T_\text{sys}},$$
and correspondingly, the reservoir's entropy changes by
$$\mathrm dS_\text{hot} = -\frac{\delta Q_H}{T_{H}}.$$
It is straight-forward to show then, that the total change in entropy of system plus environment satisfies
$$\mathrm dS = \mathrm dS_\text{hot}+\mathrm dS_\text{sys} \geq0,$$
with equality holding if and only if the system and environment exchange energy via heating when they have equal temperatures, $T_\text{sys} = T_H$.
As a consequence, in order to minimize entropy production (and, in fact zero it out completely) during this process, we want $T_\text{sys} = T_H$, and the net change in system entropy during this process can then be written as
$$\Delta S_\text{sys} = \int \frac{\delta Q_H}{T_\text{sys}} = \frac{Q_H}{T_{H}},$$
since we are assuming that the temperature of the reservoir doesn't change at all during the cycle.
Now, since the system operates on a thermodynamic cycle, and since the system entropy $S_\text{sys}$ is a state variable (state function/$dS$ is an exact differential, etc.), it must be true that
$$\mathrm dS_\text{sys,cycle}=0.$$
Therefore, there must be some other process during which the system expels an amount of energy $Q_C$ to some other reservoir via heating in such a way that the change in system entropy during this new process is the negative of the change in system entropy that we calculated before. By the same argument as above, it must be that this change in entropy is
$$\Delta S_2 = -\frac{Q_C}{T_C},$$
where $T_C$ is the temperature of the cold reservoir.
Finally, then, since system entropy is a state variable,
$$0 = \Delta S + \Delta S_2 = \frac{Q_H}{T_H}-\frac{Q_C}{T_C}.$$
Another way of looking at this equation is that the net change in entropy of the hot reservoir is negative the net change in entropy of the cold reservoir during the cycle, and hence the net change in entropy of the universe is zero during the cycle.
Efficiency and work
Now, none of this seemed related to the fact that efficiency goes to 1 as the ratio of $T_C$ to $T_H$ goes to zero. This comes in in the following way. First, the net work output during one cycle is
$$W_\text{out} = Q_H-Q_C,$$
and hence the efficiency of the engine that we've just made is
$$e = \frac{W_\text{out}}{Q_H} = 1 - \frac{T_C}{T_H},$$
after some algebra. Based on our calculation above, this must be the maximum efficiency of any engine operating between these two temperatures. However, if we change the temperatures, then we can change the efficiency. The reason the efficiency goes up as the temperature ratio goes down is that $W_\text{out}$, being the difference between the heat flows, must go up if, say, we lower $T_C$ (because then $Q_C$ goes down) or if we raise $T_H$ (because then $Q_H$ goes up).
In some sense, this part really doesn't have much to do with entropy at all, because from the thermodynamic perspective, entropy production (which is the increase in entropy of an isolated system) is a measure of how much work we could have done if we had done the process reversibly, but we have already designed the perfect engine operating between those two particular temperatures above, so entropy doesn't have anything else to say.
Best Answer
Remember that the Carnot cycle is for an ideal engine, one that operates as efficiently as possible given hot and cold bath temperatures $T_H$ and $T_C$. The only restriction that the Second Law of Thermodynamics gives is that entropy cannot decrease in an isolated system. It does not say that entropy must increase. Since the Carnot engine represents "the best you can do" in such a system, it makes sense that it would have the lowest possible allowed change in entropy (namely, zero). So the change in entropy of the reservoirs will be >0.