An easy way to "guesstimate" the value of large capacitors is to charge them to constant voltage, remove the charging supply, and connect a fixed resistor across the capacitor. The time required for the capacitor to discharge through the resistor to 37% of the charging voltage is equal to one "time constant" which is the product of the discharge resistance in ohms and the capacitance in farads.
For example: charge a 500 μF capacitor to 10 V and connect a 100 kΩ resistor across it. The product of 500 x 10^-6 F and 100 x 10^3 Ω is 50,000 x 10^-3 seconds or 50 seconds, at which time the voltage will now be 3.7 V. If you leave your POS multimeter connected while discharging through the 100 kΩ resistor, it adds a parallel load of 7.8 MegΩ, which is insignificant compared to your ability to watch the meter and operate a stop watch at the same time. If you can measure the time to within a few fifths of a second, you can calculate the capacitance in farads by dividing the time in seconds by the resistance value in ohms.
You can actually leave the resistor and voltmeter connected to the capacitor while charging the capacitor. Just remove the power supply connection to the capacitor and start your stop watch at the same time. When the voltmeter indicates 37% of the initial charging voltage stop your stop watch. That's one time constant. Divide by the resistance in ohms and you have the capacitance in farads, accurate to at least one significant figure, maybe two.
