If by instantaneous you mean what is commonly called instationary, the rules of the turn allow to trade altitude for turn rate such that the maximum load factor can be achieved even if the drag involved in doing so far exceeds the thrust available from the engines.
Contrast this with a stationary turn were no more drag may be created than can be handled by the engines such that altitude is maintained.
Then the answer is easy: The maximum bank angle $\varphi$ for a coordinated, instationary turn can be obtained from the maximum g load factor $n_z$ the plane can sustain. For small vertical speeds, use the formula for level flight:
$$\varphi = \arctan\left(\sqrt{n_z^2-1}\right)$$
If the vertical component of the flight path angle $\gamma$ should not be neglected, the load factor can become slightly higher since gravity has a forward component which does not add to the loads in the z-direction:
$$\varphi = \arctan \left (\sqrt{n_z^2-\cos\gamma}\right)$$
The answers to this question discuss the issue in more detail.
As to the time the pilot can sustain the bank angle: This is really about the load factor, and for that we have Eiband diagrams. Essentially, much depends on pilot position and the techniques used to avoid blackout. A few trained pilots when sitting can suffer through 9g for a few seconds, but a more normal limit would be 12g (equivalent to $\varphi = 85^\circ$) for 0.04 seconds, decreasing to 5 g (equivalent to $\varphi = 78^\circ$) when durations in excess of 0.2 seconds are involved.
