Including expected, but not visible nails, between 1 000 and 100 000.
The remaining uncertainty can be reduced by a [insert your favorite useless education here] student, ...
...weighting them (and a few individual nails). You don't want to wait until he counted them, right?
I DON'T KNOW. However I will try a stupid approximation technique:
Let N be the number of nails ,
V be the average volume of the nail bulk, and u be the average volume of a single nail (at different orientations). Let l be the length of a single nail. Exclude the empty spaces between the nails!
Identify the average shape of the nail bulk. Find u with respect to l (I.e. u = f(l) ). Find V with respect to l (I.e. V = g(l) ).