ICE Thermal Storage Energy Saving Calculation
The following energy saving calculation and the resulting formula: H = (C - B) / (C - D) (see bottom of this page) has been reviewed and verified by a third party expert. The resulting formula is used in a Spreadsheet to show the energy saving rate based on different average day and night time temperatures. 
This energy savings is in addition to and independent of the 
Fall/Spring/Night (Mild) energy savings which is due to the temperature difference between indoors and outdoors (without freezing water at night). However the energy saving calculations are similar.
The energy/electricity saving rate is relative to an identical (size, model, style, etc.) refrigerator and A/C that does not freeze water at night - otherwise the comparison would be partial.
I have assumed that there will always be some ice in the water/ice reservoir - its water temperature never rises above 0 oC. Therefore for the same weight of water, the same amount of heat must be extracted to freeze it during day or night - regardless of outdoor temperature. 
The Energy (electricity) Savings %  =  1 - (Wn / W
"d" stands for day and "n" stands for night.
d is the amount of work (electricity) that would have to be done during the day (hot outdoors) to freeze the water in the reservoir if it wasn't frozen during night.
Wn is the amount of work that has to be done during night (cold outdoors) to freeze the (same weight) water in the reservoir.
The amount of heat extracted from water (to freeze it) is the same because the weight and temperature of the water in the reservoir is assumed to be the same during day or night. 

As previously explained:
COP = Q / W   or   W = Q / COP
Therefore W
n / Wd = (Qn / COPn) / (Qd / COPd)
Since Q
d = Qn therefore:  Wn / Wd = COPd / COPn
COP = cold temperature (
Tcold) or the temperature of the evaporator, divided by the temperature difference between the evaporator and the condenser (Tdiff).
       COP  <=  Tcold / Tdiff
The real COP will be considerably less than 
Tcold / Tdiff depending on the model of the compressor, the type of refrigerant and other variables. To simplify let's assume (for both R/Fs) the Real
COP =  0.423
Tcold / Tdiff      
is a random number less than 1.  It can be any number.  It's the inefficiency in both R/Fs.
Spreadsheet column definitions:  
B = Average Night (5 Am) Outdoor Temperature = temperature of Window R/F's condenser at night.
C = Average Day time Temperature = the temperature of window R/F's condenser during day.
D = The temperature of the R/F's evaporator (Water/Ice Reservoir fixed at 
0 oC).
dCD = Temperature difference between day time R/F's evaporator and condenser.

nBD = Temperature difference between night time R/F's evaporator and condenser.
Substituting these variables in our formula, we get:
d =  0.423D / Td     and     COPn =  0.423D / Tn  
Or alternatively: 
n / Wd = COPd / COPn = (0.423D / Td) / ( 0.423D / Tn) =  Tn / Td
The Energy (electricity) Savings % = 1 - (Wn / Wd) = 1 - (Tn / Td) = (Td - Tn) / Td =
CD) - (BD)) / (C - D)  =  (CB) / (CD) 
All temperatures must be in Kelvin. However temperature differences (
C-B) and (C-D) can be in Celsius since one Celsius is equal to one Kelvin.
According to this formula the energy savings is directly proportional to the temperature difference between day and night (c - b) and inversely proportional to the temperature difference between average day time temperature and the evaporator temperature (the water/ice reservoir).  Therefore additive substances that can raise the freezing temperature of water (such as dust, testosterone or long chain alcohol) will increase the energy savings rate.

The heat extracted from water (at night) to freeze it is exactly the same as the heat absorbed by the ice (during day) to melt it. Therefore the greater the amount of heat (the greater the weight of water that is frozen at night and subsequently melted during day time) the greater the energy savings.