Well, I asked for peer review and I got it. And when I'm wrong, I usually want to know it early. And I was wrong.
I had asked if anyone else had any insight into the resistance of the fuses. The bottom line is that their resistance is in the low numbers of milliOhms. miimura and I just tested one. So, while I was considering changing the fuses in the charger to the same model as the one in the battery junction box, I now know that this is unnecessary and I will keep the same model even as I engineer an external fuse box for them.
miimura and I met up in his laboratory and connected an electronically controlled current sink to a 24V power supply with one of the fuses in-line. Although we don't know the accuracy, the precision of the current sink is given to the nearest 0.001 Amps.
At 3.000 Amps of sunk current, we confirmed a voltage drop of 5.6 mV across the resistor using two multimeters that agreed perfectly to within their 0.1 mV precision. We then reran the test at 6.000 Amps and observed 11.3 mV across the fuse on both multimeters. So the measurement of the resistance was highly linear, as one would expect.
V=IR AKA R=V/I = 0.0113 V / 6.000 A = 1.88 mOhms. This is similar to the other resistors that I had proposed to use so there is no advantage to using them; I'll just keep using the ones that Tesla had put in the charger.
So, how much power will be dissipated at 40 Amps? V=IR=0.00188 * 40.0 = 0.0752 volts dropped across the resistor. P=IV=40.0 A * 0.0752 V = 3.01 Watts.
The specs sheet that alflash posted gave the spec at 3.8 Watts and it could be a matter of our measurement precision or, more likely, that the specs sheet gave a maximum energy dissipation. In any case, the fuses are fine to use and I will continue to use this version going forward.
Thanks miimura and alflash for helping check my work. I'll edit the initial post to point out my error.