When I get unexpected results, I find doing the reverse maths can bring a sanity check into the situation. Might not help you find the problem, but might tell you whether you're in the ball park for what you're seeing.
4.1v to 3.2v = 0.9v drop
0.9 x 12 = 10.8v drop
You've said 500w on a 49.2v system, so close enough to a 10a draw
r = vDrop / i
r = 10.8 / 10
r = 1.08Ω for the entire pack or 90mOhms per string of cells.
To get a string, you look like you put 8 cells together in parallel. 1/R(total) = 1/R(1) + 1/R(2) ... 1/R(8).
Because I can't do maths until I get my caffeine shot. I'm going to ball park it and say 0.8Ω per cell is close enough (Round up to 100mΩ instead of 90mΩ for the pack)
That's around 800mΩ per cell
According to this thread:
"Typically a "good". 18650 cell would have a discharge IR of between 50 and 100 mohm (0.050 - 0.100 ohm)"
https://endless-sphere.com/forums/viewtopic.php?t=60364
Your cells are out by at least one order or magnitude. To give you another reference, my 16Ah multistars (For comparison to your 17Ah string of cells) are 3mΩ, or two order of magnitudes smaller.
All this maths just to show you what you already know. Nothing gained. (Not much ventured either.)
But maths can still tell us more. You said you had no drop at 1A for each cell individually.
Theoretical loss at 800mOhm internal resistance
800mOhm = vDrop / 1a
vDrop should have been 0.8v at a 1a draw. (Makes sense, you are applying 10 times the load, but 8 times the supply, and seeing 0.9v per cell drop. The difference can be accounted for by the rounding I did earlier)
So if you didn't see that, you can conclude one of three things:
1. There's a problem with your connectors/soldering.
2. You damaged the cells with heat during assembly.
3. Your 500w controller might be drawing a lot more than 500w (Not uncommon).