Water thaws at 0°C. The latent heat of fusion of water is 344,000 J/kg, which means that to thaw the frozen water inside your cubes requires much, much more energy than you're accounting for.

Steel, on the other hand, undergoes no such phase transformation, and so it cannot invoke the magic of phase transformations to cool your drinks.

Water thaws at 0°C. The latent heat of fusion of water is $L_i = 344000$ J/kg, which means that to thaw the frozen water inside your cubes requires much, much more energy than you're accounting for. Steel, on the other hand, undergoes no such phase transformation (edit: at comfortable drinking temperatures), and so it cannot invoke the magic of phase transformations to cool your drinks.

To put some numbers on it: how cold would a cube of solid steel have to be to absorb as much heat as the same volume of melting ice? We want $$ \rho_s C_s V \Delta T = \rho_i L_i V \quad \Rightarrow \Delta T = \frac{\rho_i L_i}{\rho_s C_s} = 87.3 {}^\circ \mathrm{C}. $$ So to match the performance of a melting ice cube at 0°C, your steel cube would have to start out at -87°C or so, which is clearly out of the range of household freezers.