Well, as an approximate lower bound you can use the listed amp-hours of capacity and the battery chemistry to determine how many atoms of silver would be needed. To produce one amp-hour requires 3,600 Coulombs of charge. In silver-zinc batteries, each atom of silver is reduced twice, giving off two electrons.
A single electron has a charge of 1.602×10−19 Coulombs, so you would need (1/2) / 1.602×10−19 = 3.12x1018 silver atoms, or 3.12x1018 / 6.022x1023 = 5.18x10-6 moles of silver per Coulomb.
Multiplying this by the 3,600 Coulombs needed for one amp-hour, we get a minimum of 0.019 moles of silver per amp-hour. However, one cell of silver only produces ~1.55 volts, so the real battery has 20 cells. Therefore, we need 0.373 moles per full amp-hour. As silver weighs 107.87 g/mol, this would weigh 40.2g.
Finally, we can calculate the minimum silver for each type of battery:
LM Descent: 400 Ah x 4.5 batteries = 72.5kg of silver (out of 270kg total)
LM Ascent: 296 Ah x 2 batteries = 23.8kg of silver (out of 112kg total)
CSM: Returned to Earth.
Having 20-30% of the batteries be silver makes sense, as there was also roughly the same amount of zinc in the batteries, as well as electrolyte and a casing.
The descent stages obviously stayed on the Moon, and there were 6 of them: 72.5kg x 6 = 435kg.
The 6 ascent stages (which impacted the Moon after ascent): 23.8 x 6 = 143kg.
Also, although they were not included in your question, the lunar rovers used silver-zinc batteries of 23 cells each, and the batteries of the 3 lunar rovers, at 242 amp-hours each, would add 34kg of silver total.
Total: 612kg of silver, worth ~300,000 USD today. Of course, "Moon silver" would probably fetch a higher price.