If the photons are coherent, i.e. the source is a laser, then the cone with have a minimal spread.

There will always be a cone. Even a laser has a cone and a narrow frequency band. It is not a "single" frequency, nor is it perfectly coherent. Within the confines of almost any experiment, it can be considered both, though.
This is all due to the fact that the photons carry momentum. The momentum is slight in comparison to the mass of the atom, but conservation of energy says that the atom MUST receive an equal and opposite change in momentum. This means that the emitted photon is slightly off "perfect."

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A lot depends on the "sharpness" of the tube's end, and what percentage of photons go "near" it. Photons that pass roughly 1 wavelength from a sharp edge will be diffracted. This diffraction will induce "banding" due to constructive and destructive interference.

This is easy to test at home. If you have a laser pointer and anything with a straight, opaque corner. If you set both up on stands, such that the laser beam is pointed at the corner, then the projected "shadow" will have bands of light and dark at the boundary of the spot where it is "cut off."

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If the photons had a wavelength of near 1 mm... probably even 0.5 mm or o.25 mm... then they could be diffracted into the wall and absorbed quite easily. Or they could be diffracted "through" the wall. (For reference, microwaves have wavelengths in the ballpark of centimeters. This is one reason they don't pass through the tiny holes in the shielding on your microwave oven's door.)

I'd need to actually crunch some numbers as to the spread of the cone... it would be a function of the wavelength of the photons.

FWIW, any photons diffracted into the blackbody would eventually be re-emitted in a random direction, and with a random number of photons. Random within the limits of the thermal excitations of the electrically charged atoms.

The blackbody emissions are merely the excitations of the EM fields by moving, charged particles.