Cooling a box 1X2 m with TEC module
I assume your box has a height of 1m, thus it could store 2m³ of water. I fill it with only 500L. And the water has a temperature of 25°C. I'd like to drink it cool and I prefer 15°C. This makes a difference of 10K. To remove 10K of warmth from 500L of water (4 kJ/(kg*K)) requires to extract 20MJ of heat.
This will not happen instantly like switching the peltiers on and 20MJ are gone. Instead this needs some time, I assume a day or 24h. Dividing 20MJ with 24h gives 834kJ/h. And kj/h can be converted to kW which is roughly 0.232kW or 232W for an entire day.
Finally peltiers aren't that efficient and if you load your set of peltiers with 250W they won't extract the same amount of heat. This largely depends on the thermal resistance on the hot side. A flat peltier in the desert sun experiences a high thermal resistance causing very low efficiency. If you add instead a good heat sink and maybe a fan, too you'll end up in the 20% range, thus 1W of electric power turns into 0.2W of extracted temperature power. (And also add the fan's power consumption and cost for the heat sinks. The box would also be dug into the ground in the shadows.)
Now this is where the guessing starts and I multiply 0.232kW by 5 (for the very optimistic 20% efficiency of your overall construction cooling 500L of water from 25°C to 15°C within 24h) and this yields 1.25kW of electric power each hour for an entire day. With 5V elements in parallel you'd have to supply 250A to all of them.
If you need the formulas feel free to comment, so I add it later on. I also used common »power« instead of distinguishing between scientific power and  energy.