Roughly a truncated cone with diameters ~7 nuts and ~9 nuts, and the cup is ~12 nuts high (quite hard to tell due perspective and nits of different sizes). Throw in an extra layer to account for the heap at the top (which is a dome, but the dome is taller than 1 hazelnut, so treating it as a shorter layer should give some error cancellation) to give a height of 13. The volume using the formula for a truncated cone of those dimensions is ~657 cubic hazelnut diameters. Random sphere packing is 64% (though wall effects should decrease this number) giving a total of 420 nuts (nice).

Answer

This ends up being about 5% lower than the true answer. I’m surprised it’s that close. The underestimate makes sense given the wall effects I mentioned, but the error bars on my eyeballed measurements are surely higher.