Answer
No, an insignificant amount of heat will be exchanged, so the water temperature drop is minuscule.
Work Step by Step
The temperature of the water in the swimming pool is somewhere in the twenties $(\mathrm{C}^{\mathrm{o}}).$ The mass of the water is some orders of magnitude greater than the ice cube's. The water would have to be in equilibrium with the ice to drop to $0^{\mathrm{o}}\mathrm{C}$ temperature.
In order to achieve this, it would have to pass on an enormous amount of heat to the ice cube. The ice cube, however, has a small mass and can take only a small amount of heat (of fusion), enough to melt. The water's temperature does not significantly fall, as it passes only a small amount of heat.
So, no, they are not in equilibrium.
