General principles
You should probably just run the experiment over the weekend, but here are the relevant bits of physics:
Typically, saltwater freezes at a lower temperature, so you will probably reach a point where your freezer no longer fully changes the state of the water to solid. You probably want the freezing to happen no matter what.
The reason that you probably want freezing no matter what is because the colder the bottle gets, the faster it will exchange heat with the aquarium. The heat flow will be proportional to the difference in temperatures. Ice has this really nice property that it forms and melts all at one temperature, so you do not have to increase the rate at which cold is pouring out of the system in order to store more coldness in the system. That's a big win.
Whether the water should be salted or not depends heavily on the temperature of your freezer, among other things. Water has about twice the specific heat of ice, so that suggests that you want as much water-stage as possible. So you have two competing things here: first off that the heat transfer is greater for a lower freezing temperature, pushing the freezing temperature up; second, that the water holds more heat than the ice will, pushing the freezing temperature down.
Calculating the maximum time.
The model that you want is something like: from temperature $T_0$ to $T_f$, we have ice; then from temperature $T_f$ to $0~^\circ\text{C}$ we have liquid, then we assume that the water and saltwater will heat up the same amount from then on. Your aquarium is at some temperature $T_1$ and so the heat exchange will go like $\frac{dq}{dt} = k (T_1 - T)$ for some unknown $k$, leading to temperature changes due to $q = c m T$ where $m$ is the mass and $c$ is the specific heat capacity per unit mass. Curves are then looking like $T_\tau(t) = T_1 - (T_1 - \tau) e^{-kt/(cm)}.$
From $T_0$ to $T_f$ requires a time $t_0$ such that $T_{T_0}(t_0) = T_f;$ solving this yields: $$ t_0 = \frac{m ~ c_\text{ice}}{k} ~\log\left(\frac{T_1 - T_0} {T_1 - T_f}\right).$$ Then there is some heat of fusion $h_f$ with timescale $t_f$ given simply by $ m ~h_f = k (T_1 - T_f) t_f,$ so that:$$t_f = \frac{m~h_f}{k (T_1 - T_f)}.$$Finally, heating up to $0~^\circ C$ takes a time$$t_1 = \frac{m ~ c_\text{water}}{k} ~\log\left(\frac{T_1 - T_f} {T_1 - 0~^\circ C}\right).$$The total time is $t^* = t_0 + t_f + t_1$ which is maximized independently of $m/k$ by setting its derivative to zero,$$\frac{k}{m} \frac{dt^*}{dT_f} = \frac {c_\text{ice}} {T_1 - T_f} + \frac{h_f}{(T_1 - T_f)^2} - \frac {c_\text{water}} {T_1 - T_f} = 0$$This is solved by $T_f = T_1 - h_f / (c_\text{water} - c_\text{ice}).$
Now of course the addition of salt will change these parameters, but it's important to see the order of magnitude of what we're talking about here. This critical temperature $T^* = h_f / (c_\text{water} - c_\text{ice})$ is something like (300 kJ/kg) / (200 kJ (kg °C)) ~= 1.5 °C: it is tiny in comparison to $T_1$.
This means that the extra heat capacity gotten from shifting the temperature at which the water freezes is tiny in comparison to the higher heat transfer: you want to push this temperature as high as possible rather than as low as possible. So you want no salt.
What else can you do?
The absolute simplest way to make the ice last longer is to lower the parameter $k$ above. The easiest way to do this is to get some air between the ice and the water. Those foam pads that one wraps around beers might work best for this, but they might also pollute the water and/or be eaten slowly by fish. You might instead try a zip-lock bag or even a combination of the two. Of course, then you will need to weight the bag down so that it stays in the water.