How much g-force is experienced in a 45° turn?
The formula for the load factor $n_z$ in a turn is

$$n_z = \frac{1}{cos\phi}$$

where $\phi$ is the bank angle. This makes the load factor in a 45° turn 1.414 or $\sqrt{2}$.



Since Michael correctly mentions the simplifications in his comment, let me expand the formula above. In case of a climbing turn, the load factor depends on the coordinate system used. If it is measured in airframe-fixed coordinates, the load factor will be split into a z-component ($n_z$, pointing down) and an x-component ($n_x$, pointing fore- or backward, depending on the aircraft's pitch attitude $\Theta$; nose-up is positive):
$$n_x = -\frac{sin\Theta}{cos\phi}$$
$$n_z = \frac{cos\Theta}{cos\phi}$$

For an uncoordinated turn the value needs to be modified further with the relationship between bank angle change for a given sideslip angle, which is aircraft-specific.