The power is the rate you need to add energy. In each time interval $\Delta t$, you have to accelerate a new slug of water $\Delta m$ from zero up to speed $v$, which requires energy $\Delta E=\frac{1}{2}\Delta m v^2$. Now, $\Delta m=\rho A v \Delta t$, so $$\frac{\Delta E}{\Delta t}=\frac{1}{2}\rho A v^3.$$

The power is the rate you need to add energy. In each time interval $\Delta t$, you have to accelerate a new slug of water $\Delta m$ from zero up to speed $v$, which requires energy $\Delta E=\frac{1}{2}\Delta m v^2$. Now, $\Delta m=\rho A v \Delta t$, so $$\frac{\Delta E}{\Delta t}=\frac{1}{2}\rho A v^3.$$

As to where in your derivation you went wrong, it is a misapplication of $P=F\cdot v$. This expression is valid if $v$ is constant during the period when $F$ is applied. That is not the case here: $F$ accelerates the differential mass of water $dm$ from 0 to $v$; its average speed during this process is $v/2$.