Why does the faster-moving air over the static port not result in a lower static pressure?
Bernoulli's principle would also imply that the pressure observed from the static port would similarly decrease as the aircraft accelerates.


It's not that simple. Aircraft are complicated shapes. The fuselage is composed of all sorts of angles of attack, things sticking out into the airflow, and the pressure distribution of air across it is very uneven, just like that of a wing is. Take a look at this diagram from an FAA mechanics publication. 

There are zones (like points 2,3, and 5 in the diagram) where the static pressure deviation is zero. This is where the engineers put the static ports.

Another way to think about this is that you're right that the airflow over the static ports should make the pressure lower--- but the question is: lower than what? The answer is: lower than the airflow's stagnation pressure, when the outside airflow is at rest (relative to the aircraft). Which happens at the nosecone, or the leading edge of the wing... or at the pitot tube. It's lower than that by the same amount that the pitot pressure is higher than ambient, making the pressure at the static port exactly ambient outside pressure. Which should make sense, because in the reference frame of the air, that static air isn't doing anything but sitting around waiting for an airplane to come by. 

The Bernoulli effect does, however, have an overall average result: the cabin pressure in the airplane is lower than static, because the angle of attack of the various surfaces of the fuselage generates suction, pulling a small amount of air out of the cabin, just as it does for the wing if there are little holes in the wing. Which is why you get the little blip when you pull the Alternate Static open.