why is the current of a aluminum-hydrogen galvanic cell low
My chemistry isn't so good but I expected that you would be hoping for a 1 to 2 volt cell. You don't give any details of the size of your cell but I imagine that it's small or that the surface area of the electrodes may be small and that's limiting the maximum number of electrons it can produce. 

All cells have internal resistance which causes the voltage to droop in proportion to the current drawn. We can model a cell (or battery) as a "perfect" voltage source of potential equal to the open-circuit voltage with a series-equivalent-resistor. By measuring the open-circuit voltage and making a test a known load and we can calculate the equivalent internal series resistance.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1 and 2. Open-circuit and loaded voltage measurements.


For the open-circuit measurement we use a voltmeter with a high input impedance so that negligible current is drawn from the cell. You got 1.34 V.
We take a second reading with the 47 Ω load. You got 0.07 V.
The voltage drop across the internal resistance is then 1.34 - 0.07 = 1.27 V.
Let's double-check your current reading by calculating the current through RL, 47 Ω, using Ohm's Law. \$ I = \frac {V}{R} = \frac {0.07}{47} = 0.00149~A = 1.49~mA \$ which is very close to your reading. Excellent.
We can now calculate \$ R_S = \frac{V}{I} = \frac {1.27}{0.00149} = 852~Ω \$.


With that calculation we can see that if we loaded the cell with an external 852 Ω resistor we would only get half voltage, 0.67 V, at the terminals. If you use a lower resistor (bigger load) the voltage will drop further.

Maximum power

If you leave your cell open-circuit you can get maximum voltage out of it. This isn't much use as the current is zero and since \$ P = V \cdot I \$ the power out is zero.

If you short circuit the battery you can get maximum current out of it but with zero volts across the short you get zero power again. (All the power is dissipated in the internal resistance.



Figure 3. Power transfer graph. Source: Wikipedia Commons. (See link below.)

The maximum power transfer theorem shows that the most power you can get from the cell is when \$ R_L = R_S \$.



Other info

The ampere is defined as "a unit of electric current equal to a flow of one coulomb per second".

A coulomb is equivalent to the charge of approximately \$ 6.242 × 10^{18} \$ electrons.

You're getting 1.5 mA so that's \$ 6.242 × 10^{18} × 0.0015 = 9.36 × 10^{15} \$ electrons per second. Congratulations!