As one litre of ethanol weighs about 790 g, you want 15 percent of those grams to be ammonia, that is 790 x 0,15 = 118,5 g, which is 118,5 / 17 g/molNH3 = 7 moles roughly.
NH4Cl + NaOH -> NH3 + H2O + NaCl
You will need to add 7 moles of NH4Cl and 7 moles of NaOH.
Hence:
NH4Cl: 7 x 52,5 g/mol = 367 g.
NaOH: 7 x 40,0 g/mol = 280 g.
CaO: Enough to absorb 7 moles of water (CaO + H2O -> Ca(OH)2).
I'm quite positive that a solution of 671,5 g ethanol and 118,5 g of ammonia does not have the volume of 1 litre, so you might have to make a slight excess just to be sure you get enough. Now that you get the idea I'm sure it won't be a problem though...