This is a great question, and the answer relates intimately to why turbofan engines equipped with afterburners require variable geometry exhaust nozzles. Without increasing the throat area to accommodate the larger volumetric flow rate, lighting the afterburner would back-pressure the fan and very possibly lead to a compressor stall. Similarly, mechanically reducing the downstream area (all other things being equal) will require the flow to have a higher upstream stagnation pressure, which means the fan/pump will be required to work harder.
Now, as to your question about why the flowrate decreases when the exit area is closed, we need to expound a bit on how fans and compressors operate. The fan speed, massflow rate, and pressure ratio are related in a complex way and are usually represented graphically by a fan map.
For a given rotational speed, there is a single steady-state characteristic relating pressure ratio and massflow rate. The shape of this curve can vary (e.g. compare the 40% Nf line with the 100% Nf line above), but generally speaking the higher the pressure ratio, the lower the massflow rate for a given engine RPM. This makes some intuitive sense because the faster the bladetip velocity compared to the axial velocity, the higher will be the flow turning within the bladerow. 568
Work done and pressure rise are proportional to flow turning within the rotor, so higher pressure ratios are positively correlated with lower massflow rates/axial velocities (up to a point).
To truly understand the causal relationship between massflow rate and back-pressure requires that we abandon steady-state thinking altogether. If the exit area is reduced, unsteady compression waves propagate upstream at the speed of sound, incrementally increasing the static pressure at the exit of the fan. This increased back-pressure means that the entering flow now encounters an adverse streamwise pressure gradient and slows down. This slower flow is then worked harder by the spinning bladerow, which results in larger stagnation pressure and temperature rises.
Remember that the flow always exits the device at atmospheric pressure so long as it is subsonic, precisely because of the information propagated upstream by unsteady pressure waves. Thus, if reducing the exit area means a higher exit Mach number is required to conserve mass, the total-to-static pressure ratio must increase. A fixed exit static pressure and increased total-to-static pressure ratio demands that the upstream stagnation pressure increase, and so the upstream turbomachinery will be affected.
If you are looking to put numbers on things, the isentropic flow function is a useful and straightforward way to determine the massflow rate of a compressible fluid if other of the fluid's basic properties are known. In general, the massflow rate of a fluid through a cross-sectional area $A$ is equal to
$\dot{m}=\rho VA$.
Now, if the fluid is compressible and the Ideal Gas Law applies, then
$\dot{m}=\rho VA=\left(\frac{P}{RT}\right)(M\sqrt{\gamma RT})A=PAM\sqrt{\frac{\gamma}{RT}}$.
Both the stagnation temperature and stagnation pressure are preferred flow variables to their static counterparts, so the above equation can be rewritten as
$\dot{m}=P_0 \left(\frac{P}{P_0}\right)AM\sqrt{\frac{\gamma (T_0/T)}{R(T_0)}}$,
and the stagnation properties (as well as the through-flow area) can be moved to the LHS of the equation:
$\frac{\dot{m}\sqrt{T_0}}{P_0 A}=\left(\frac{P}{P_0}\right)M\sqrt{\frac{\gamma}{R}\left(\frac{T_0}{T}\right)}$
If the flow is isentropic (as we are assuming), we know that
$\frac{P}{P_0}=\left(\frac{P_0}{P}\right)^{-1}=\left(\frac{T_0}{T}\right)^\frac{\gamma}{1-\gamma}$,
which gives us
$\frac{\dot{m}\sqrt{T_0}}{P_0 A}=M\sqrt{\frac{\gamma}{R}}\left(\frac{T_0}{T}\right)^{\frac{1}{2}+\frac{\gamma}{1-\gamma}}=M\sqrt{\frac{\gamma}{R}}\left(\frac{T_0}{T}\right)^{\frac{1+\gamma}{2(1-\gamma)}}$. 568