I wanted to weigh in on this discussion, if only to sharpen my own understanding of things.
First, the Wehnelt voltage. I don't think this should really be subtracted from the initial HV. I think the Wehnelt determines how many electrons emerge from the gun and from which part of the crystal/tungsten filament, but not their final energy. Typically, the Wehnelt voltage is relative to the emitter, so whereas the electrons emerging from the filament see a decelerating field on their way towards the Wehnelt, they are accelerated again by the same amount after leaving the Wehnelt aperture. I think of it this way: When you consider the whole gun module as a "closed system" ensemble which is at voltage E0, then electrons coming out of it will have that potential, regardless of what's happening inside that ensemble.
I tried to do some bookkeeping of all the energy deltas. Here's my list, feel free to comment.
First, the E0 itself (as set by the microscope) will likely have some (systematic) "setpoint error", for example because the voltage entering the HV cascade is determined by some DAC with a finite number of steps. The "problem" here is that this will then be a fixed percentage of the "full range", so if your HV subsystem can go up to 30kV and uses a 10-bit DAC to set the value, you're looking at a discretization error of ~30V just from this. The HV supply should also specify a certain maximum "ripple", which is typically an order of magnitude better than this (because it would otherwise give bad ("fluctuating") imaging results).
There may be a small voltage that needs to be subtracted from the supplied HV because that's applied to the filament itself to heat it up, but that would be just a few volts.
Then, there will be a spread in the energy with which the electrons escape from the filament or crystal. This is worst for tungsten (because you typically heat it up the most) and best for a cold FEG. Regardless, it will also be only a few eV, so I don't think this should worry us. This is one of the causes of chromatic aberration which is why microscope builders try to keep it as small as possible, or even insert monochromators for ultimate imaging performance (at the cost of intensity, so I suppose nobody here would use them unless you're also doing EELS).
Then, electrons can also undergo further ("perpendicular") acceleration inside the scanning column, so that increases their energy as a function of their final distance from the optical center. In case you're scanning a big field of view, this may be significant. Speaking for the microscope my team is producing, we're scanning electrostatically with a few hundred volts on the scanning poles, so my guess is that electrons aimed for the edges of the maximum field of view have their energy increased by the same amount (and worse: that means the electron energy varies over the image - interesting when doing mapping!)
Finally, I think in many cases there will be some charging happening at the sample surface. This is the hardest to quantify. But unless you're analyzing some piece of perfectly grounded metal, I'm sure there will be some equilibrium between the electrons you're blasting onto a single tiny point on the sample and them flowing away to earth again.
And when the X-rays are finally being created in the sample, there is (as already mentioned here) the gaussian spread in the detector itself (and the resulting "convolution" of the measurement) and I think that should not be underestimated. After all, even for a perfect SDD with a MnKa FWHM of 119 eV, the FWHM of a 15kV peak would be something like 190eV. So, I think that even if "perfect" 15000.0eV electrons are hitting your specimen, you should not be surprised to find some signal in your spectra a few hundred eV higher than that.
By the way, this is all not to say that I don't find the whole bremsstrahlung phenomenon super weird. Since we are seeing photons with energies up to E0 in our spectra, that means that these were emitted by electrons losing all their energy during deceleration (so far, I'm nodding) but since each photon corresponds to a single quantum mechanical event, that must mean that some electrons come to an abrupt stop from full speed in one infinitesimal event. How?!
Wasn't it Bohr himself who said "Anyone who is not shocked by quantum theory has not understood it" - I'm definitely still in the "shocked" phase.