Is there a relationship between control surface deflection and a particular turn radius or bank angle?
You need the ailerons only to bank the aircraft; once the bank angle is reached, the aileron deflection is mostly close to zero. I assume we are talking about stationary turns, so speed and aircraft mass will not change significantly over the duration of the turn.

While turning, several moments need to be balanced to keep the roll angle constant, and a good design does this without requiring aileron input:


The rotation around the vertical axis causes more airspeed over the outer wing, increasing its lift. This causes a rolling moment into the turn.
The rotation around the vertical axis causes a sideslip condition at the vertical tail which causes a yawing moment against the turn.
The inertial forces try to pull the wings level. This causes a rolling moment against the turn.


However, the elevator needs to be held at a slightly more negative angle than in level flight to pull the required load factor $n_z$. In a turn, the load factor is proportional to the bank angle $\Phi$:
$$n_z = \frac{1}{cos \,\Phi}$$
Radius $R$: $$R = \frac{v^2}{g\cdot tan\,\Phi}$$
Angular velocity $\Omega$ (rad/sec): $$\Omega = \frac{v}{R} = \frac{g\cdot tan\,\Phi}{v}$$
The amount of elevator deflection needed depends on the stability margin (expressed as $\frac{c_{m\alpha}}{c_{L\alpha}}$) of the airplane, its pitch damping (expressed as $c_{mq}$) and elevator effectiveness (expressed as $c_{m\eta_H}$).