You're right. You don't know. Hint: if his description is accurate, it's an open channel of constant cross section, therefore constant flow speed and constant static pressure. Which has nothing to do with your drawing or Bernoulli's principle.

A port placed orthogonally to the gutter won't be exposed to dynamic pressure, the only force it will "see" is atmospheric pressure, plus the weight of the water column directly above the port. Regardless of flow rate. He could attach a tube between the test port and the bottom of a glass burette and he would have a real-time indication of level.

Flow would only matter if the test port was oriented directly into the flow vector. If you did that, the total pressure would be the sum of static pressure plus dynamic pressure (flow). Try this picture. The only difference is that the flow path is an enclosed tube, not an open channel:

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