Actually once the manhole cover is off, you would calculate the velocity based on the height of the water column and the flow based on that velocity and the diameter of the manhole (typically 24" or 36" for a storm drain.)
I found a list of tabulated velocity heads (
http://www.engineeringtoolbox.com/static-pressure-head-d_610.html) here which would suggest a water column of 4 feet would be a velocity head of about 16 feet per second. Once it's clear of the confines of the manhole, it's mostly velocity head carrying the water but there will be some pressure head element as well. It would be fairly difficult to calculate how much pressure head to subtract in an open water column.
Assuming you have a 24" manhole, the cross sectional area is 3.14 square feet (I can do that math in my head, a = pi R^2) so the flow is then 3.14*16 = 49.6 say 50 cubic feet per second assuming it's all velocity head. That works out to 22,400 gallons per minute.
That's a lot of water and it's probably not nearly that much. If you could manage to stick a pitot gauge calibrated for water velocity at the top of the manhole, you could measure the actual velocity and then calculate the actual flow.