I don't think an f/4 Newtonian would be a good choice for photographing the moon. In fact, f/8 might be a stretch in terms of covering a sensor as large as the IMX183 (for the diagonal, which is approximately 15mm, but I'll need just the center portion of the full frame -- see below).
According to a post by Jon Isaacs the coma free field (diameter) for Newtonian reflector can be computed as follows:
However, I don't know what his criteria is for "coma free."
So, at f/4 we have: 0.022 x 4^3 ≈ 1.4mm
And at f/8 we have: 0.022 x 8^3 ≈ 11.2mm
This compares to the height of the IMX183 of 8.8mm and thus f/8 should be "safe" as long as the moon remains centered in the field.
I'm not sure whether my 6" f/6 Newtonian can cover the entire field of the IMX178 and the calculations suggest:
0.022 x 6^3 ≈ 4.7mm
And the size of the IMX178 is 7.4mm x 5mm with a diagonal of 8.9mm.
As far as I can tell, the criteria used for "coma free" in these posts is referring to visual observation and means that the size of the coma is less than the size of the Airy Disk, such that it would be undetectable in visual observation. However, I would think that the criterion would be even stricter for imaging, because we are typically sampling several pixels across the FWHM of an Airy Disk, and the full size of the Airy Disk is much larger than this. Also, I've come across other formulas on CN that have stricter values than the 0.022 multiplier. But still this is geared towards visual observation.
I also came across a formula in one of Thierry Legault's books that gives the length of the coma tail as the following:
Where "d" is the distance off-axis, and F is the focal ratio. I haven't been able to confirm this formula elsewhere however, and I'm not sure why the focal ratio is squared in this case, yet in other posts on CN there is always mention of how coma increases as the cube of the F ratio.
In any event, if we use this formula and take the off axis value to be 4.4mm (half that height of the IMX183 sensor), then the length of the coma in an f/4 Newt should be 52um, which means it would be spread across 21 pixels! In an f/8 Newt, this value drops to 13um, which would mean it's spread across 5 pixels. This still seems like quite a lot to me, although for an f/8 scope, the spot size of the Airy Disk is only ~11um, so it's not that much larger than the Airy Disk, which kind of matches up with those posts you are referring to. Although the spot size of the FWHM is only ~5um, and since we often sample for this value, an aberration of 13um would certainly have a negative influence.
Of course, a coma corrector can be used......but this locks you into the prime focal length, as it's difficult, if not impossible, to combine a coma corrector with barlows.