You have exact equations for the solution in the related question Time it takes for temperature change. Here I would add a few comments.
It is actually easier if container is thick! Then suppose that all water is at same temperature $T$ and all the air in the frizer is at the same temperature $T_e$. $T_e$ is constant.
If that is so, you can use only Fourier's law to describe how heat $Q$ leaves the container
$$\frac{\text{d}Q}{\text{d}t} = \frac{\lambda A}{d} (T-T_e).$$
$d$ is thickness, $A$ area and $\lambda$ thermal conductivity of the container.
Knowing that water cools as heat is leaving the container
$$\text{d}Q = m c \text{d}T,$$
where $m$ is mass and $c$ is specific heat capacity of the liquid, you get rather simple differential equation
$$m c \frac{\text{d}T}{\text{d}t} = \frac{\lambda A}{d} (T-T_e),$$
$$\frac{\text{d}T}{(T-T_e)} = \frac{\lambda A}{d m c } \text{d}t = K \text{d}t,$$
which has exponential solution:
$$K t = \ln \left(\frac{T-T_e}{T_0-T_e}\right),$$
$$T = T_e + (T_0-T_e) e^{-Kt}.$$
No, the temperature of the water is not that important for the performance of an evaporative cooler. This is basically because the energy needed to increase the temperature of liquid water (its specific heat capacity) is very small compared to the energy needed to evaporate the same amount of water (its enthalpy of evaporation).
At room temperature the specific heat of liquid water is 4.18 J/(g$\cdot$K) while the ethalpy of evaporation is 44.0 kJ/mol. Since the molar mass of water is roughly 18 g/mol, this means that approximately 585 times as much energy is needed to evaporate an amount of water as to increase the temperature of the same amount of water by 1 K. This means that even if the water starts at freezing temperature, is heated to 40 $^\circ$C (104 $^\circ$F) and then evaporates, less than 7% of the energy absorbed by it is used for increasing the temperature.
The temperature of the water might effect the rate at which evaporation occurs, but I guess evaporative coolers are designed to achieve 100% relative humidity in the wet bulb regardless of the temperature of the water, so this probably doesn't affect the performance.
Best Answer
It sounds like you are basically asking about how water boiling kettles often get coated with calcium carbonate when they are used with mineral-rich water. It turns out that the calcium carbonate is actually less soluble in water at higher temperatures, and so it tends to precipitate at higher temperatures.
The minerals in the water will always be left behind. So the strict answer to your question is that the same amount of minerals will be left behind regardless of the temperature of the water or how quickly it evaporates. However, I think that the way that the calcium carbonate is deposited might depend on the temperature or the roughness of the surface.
Crystals tend to form on rough surfaces, so, if you heated the water, causing the calcium carbonate to evaporate, then it would likely to form crystals on the rough surfaces. If you let the solution evaporate at low temperature, then it would leave behind the calcium carbonate on all surfaces.
This question has a good description of the situation with the kettle: https://chemistry.stackexchange.com/questions/4094/white-powder-observed-after-boiling-water-in-electric-kettle-for-many-weeks