Empirically - It is harder to stall in a steeply banked turn.

Just don't expect all aircraft to be stall proof at 60 degrees. My
Cirrus would bite anyone silly enough to try it...

It depends on the design, but most conventional (as opposed to all
moving) tail designs will reach a load and aerodynamic situation at some
bank angle where they will not stall.

My attempt to explain : -

Load on tail increases with speed. This is a function of the stability
requirements.
As mainplane AOA increases so the tailplane AOA decreases. Limiting case
here is that once the glider is fully stalled the the tailplane will now
produce an up load - pitching the nose down.
Maximum deflection of elevator gives some fixed AOA relative to the
aircraft centerline.
In any attitude where the wings are being subjected to 2G the aircraft
must be describing a "pull up" path (circular if the load is constant)

This implies the mainplane is at a relatively large AOA, say X degrees
larger than 1G. Trigonometry says the maximum AOA of the
tailplane+elevator is reduced by this amount. (They are both exposed to
roughly the same relative wind, although downwash from the mainplane can
change the angle at the tail slightly)

So - assume your wing needs to be at 4 degrees above the 1G point to
generate 2G, then your effective elevator range at 2G is the same as if
you limited your tailplane+elevator to 4 Degrees less.
Add to this that the load needed to overcome the nose down torque will
be higher than at 1G, because the CG is in front of the AC and AC moves
backwards as AOA increases (until the stall).
4 degrees is a significant fraction of the total AOA range of the
tailplane which will only operate effectively up to around 16 degrees.
You have reduced your control authority by around 25%.
Add to the fact that the airflow will probably be more turbulent and the
tailplane therefore less efficient and you have an explanation.

Sound right?