If you fill a glass with 40% ethanol solution (mol. formula C2H5OH) and leave it overnight, the ethanol concentration in the glass will still be 40%. Why does this occur?/Why doesn't any of the alcohol get evaporated?

Sorry to answer a question with a question.
You stated that the concentration stays the same overnight. But nothing was stated about the total mass that remains in the glass.
Is it then possible that the fraction evaporates at room temp is the same?
.4* evaporation rate of pure EtOH = 0.6 * evaporation rate of pure water.

I don't have my CRC handbook nor Lange's to look up their vapor pressures at 20 C

Given that ethanol has a higher vapor pressure than water, it evaporates faster. So wouldn't any ethanol-water mixture lose relatively more molecules of ethanol than water, becoming increasingly more dilute with time?

Side question: Are the volumes additive?

Raoult's law states that the vapor above a solution will reflect the vapor pressures of the components proportional to their concentrations.
Vapor press mm Hg
Temp C H2O EtOH
15 12.79 32.2
20 17.54 43.89
25 23.76 59.02
30 31.8 78.47

At the 60:40 mix you are correct that more EtOH than water will be evaporated At 71.5% water and 29.5% EtOH the proportional vapor pressures are nearly equal at 20-25C

The other question about volume additions.
No If you add 1 liter of water to 1 liter of EtOH you'll not likely get 2 liters of mixture.
Close but not quite Ideal. The subject of Partial Molar Volumes gets quite a few hits off the internet.