Provided you know the capacity of a fan (flow rate) at constant speed and at sea level, is there an analytical way to predict what the flow rate would be at altitude? Or is this specific to the fan's design?
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Following answer is speculative.
Flow rate of air ($Q$) is determined once fan's geometry, its angular speed ($\omega$), and thermodynamic state of air (in particular its $\rho,\mu$) is specified. Since geometry of fan is not being changed, let us take any linear dimension associated with it (say, length of fan blade) as a length scale, $d$. Flow rate $Q$ is determined by these variables means that there exists a functional relationship:
$f(Q,\rho,\mu,d,\omega)=$constant
which results in dimensionless groups:
$g(\frac{Q}{\omega d^3},\frac{\omega d^2}{\nu})=$constant
or
$\frac{Q}{\omega d^3}=h(\frac{\omega d^2}{\nu})$
where $f,g,h$ are functions. Now since flow Reynolds number is high (and therefore flow is turbulent), viscosity plays little role in determining the flow (hypothesis). In that case,
$\frac{Q}{\omega d^3}\approx $constant.
Since angular speed of the fan isn't varying either I would guess that flow rate of the fan shall remain constant (to a good approximation).
One may object that density of air hasn't appeared in the final conclusion, so if one were to take the fan to such a height where there is practically no air, the equation still predicts the same flow rate, which is wrong. However in this case continuum approximation breaks down, and the entire analysis would become inapplicable.
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For a fixed speed, a fan, blower or any turbo-machine in general will deliver the same volumetric flow regardless of the ambient pressure since the machine essentially scoops out a volume of air as each blade of the machine passes the machine's inlet. $$Q_{SL}=Q_{alt}$$ where $SL$ designates 'Sea Level' as reference and $alt$ as some higher altitude
But at higher altitudes there are fewer molecules per unit volume (lower gas density) and so the mass flow rate is lower with increasing altitude and barometric pressure. $$\dot{m}_{alt} < \dot{m}_{SL}$$ So since the volumetric flow rates are the same then $${\dot{m}_{alt}\over{\rho}_{alt}} = {\dot{m}_{SL}\over{\rho}_{SL}}$$ and $$\dot{m}_{alt}={{\rho}_{alt}\over {\rho}_{SL}}\dot{m}_{SL}$$ But if we were to measure these mass flows as volumetric flows relative to sea level then $${{\dot{m}_{alt}}\over {{\rho}_{SL}}}={{{\rho}_{alt}}\over {{\rho}_{SL}}}{{\dot{m}_{SL}}\over {{\rho}_{SL}}}$$ which becomes $$Q_{Malt}={{{\rho}_{alt}}\over {{\rho}_{SL}}}Q_{MSL}$$ And the $Q_M$'s are the measured volumetric flows at altitude and sea level respectively. This result shows the the measured volumetric flow is reduced as altitude increases by the ratio of air density as it decreases. And this is consistent with zero volumetric flow as one moves out of the atmosphere and no more scooping is possible.
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