Well the description is a bit unclear there.
With electrostatic discharges you get lots of both instantaneous current and voltage but little electric charge. That limits the time duration during which the current can pass and limits the amount of damage that can occur.
Over time, the current is indeed low, but the point that needs consideration here is that the current basically goes through to stages: The part where you have current and the part where you don't have current.
The part during which you have current lasts for only a short time and during that time, the current is result of the voltage and the resistance of air (which is pretty complex as air has non-linear resistance). Over time the current decreases as the electrostatic charge is depleted and the resistance of air changes due to air movement. The resistance of a volume of air through which the current is passing tends to decrease over time, but that air heats up and expands and moves away from the source of discharge meaning that the total resistance increases because the length of the conductor is increasing. This lasts for a very short time. At one point you reach the part where the resistance is too high to maintain the arc (or alternatively you reach the point at which the charge has been depleted) and then the arc breaks. From that moment on, you don't have any current.
Another point is electrocution. For that you need not only sufficient voltage but also sufficient energy. An electric outlet at say 220 V can provide "large" current for very long time (compared to how long the arc lasts) and that allows large enough transfer of energy which is expanded to damage tissue. That energy doesn't exist in case of usual electrostatic discharge.
How electrostatic discharge works can be seen in this simulation. Notice the time on the lower right part of the black screen and click on the switch and see how quickly the capacitor discharges. Something like that happens with electrostatic discharge too.