I'm puzzled. My copy of rjeloop3 suggests the Q will be about 200 at
60kHz with a 9mm wire diameter, and you'll see about 2kohms when it's
resonated. Are you not taking the output across the ends of the loop
(across the capacitor)? And with a skin depth of about 0.01" at 60kHz
in copper, certainly 3" diameter soft copper pipe would have the lower
resistance. You might have some trouble finding soft copper pipe,
though. But even hard copper pipe should have a low RF resistance.
"Reference Data for Radio Engineers" (or "Reference Data for Engineers"
in newer incarnations) has lots of good info for figuring out things
like RF resistance of copper wire. I assume your welding cable doesn't
have strands that are insulated from each other like Litz wire.

Consider that Q is energy stored divided by energy dissipated per
radian (1/2pi of a cycle). Then the net Q will be 1/(1/Q(inductor) +
1/Q(capacitor)). So if the cap and inductor have the same Q, the net Q
will be half that. And if you put a resistive load across the
coil+cap, that will dissipate power and lower the Q further.

Cheers,
Tom