This is a more complicated question than you think.
Superficially, if you have a volumetric flow rate of $V$ through a pipe with constant area $A$, then the average velocity of the fluid is given by
$$v = \frac{V}{A}$$
If you use cubic feet per second and feet squared, the result will be in feet per second.
HOWEVER
The actual velocity of a fluid in a pipe is a function of the position in the pipe. For laminar flow, the velocity follows a parabolic profile, with the points near the wall being almost stationary, and the points in the middle going fastest. In that case, there is no direct answer "how quickly will the water come out the other end" - because some water will come out much more quickly than the rest. Note - if you have a river, the flow is unlikely to be laminar (and the cross section is neither circular nor constant) so this analysis does not apply. But when I initially wrote this answer, we had no inkling that you were asking about flow in a river...
How much faster is the fastest liquid? Well, if you assume a circular cross section and a parabolic profile, then we can write the velocity as a function of radius r (for a pipe with radius R):
$$v(r) = v_0\left(1 - \left(\frac{r}{R}\right)^2\right)$$
where $v_0$ is the maximum flow velocity (at the center of the pipe).
Integrating the flow rate over the entire cross section, we find
$$V = \int_0^R 2\pi r v(r) dr\\
=2\pi v_0\int_0^R r\ \left(1 - \left(\frac{r}{R}\right)^2\right) dr\\
= 2\pi v_0 (\frac12 R^2 - \frac14 R^2)\\
= \frac12 \pi R^2 v_0$$
If we put $$A=\pi R^2$$
Then we find
$$v_{av} = \frac12 v_0$$
In other words, the mean flow velocity is half the peak velocity. Worth thinking about when you try to figure out what you are really trying to ask.