Somewhat surprisingly, focal length has very little effect at these magnification factors. DOF is almost entirely determined by effective aperture and magnification factor once you go over 1.0.
... but of course reality can be a bit more complicated. In order to compute your DOF correctly in the macro domain it is critical that you know the pupil ratio of your lens.
Short version: effective aperture is N*(1 + M), where M is your magnification factor. So, at 2.6x magnification with your lens set to f/8, you actually end up with 8*(1+2.6) = f/28.8. This means that not only will your exposure be rather long (small effective aperture letting through less light), but you will also see significant diffraction softening (for most current APS-C sensors of around 16-24 megapixels the small photosite size implies that sharpness starts to decrease when the effective f-number exceeds f/8). You could open up the aperture to counter this, but that necessarily decreases your DOF again.
Longer version: effective aperture is N*(1 + M / R), where R is the pupil ratio (of course, you have to use the right "polarity" of the pupil ratio here, i.e., entry pupil / exit pupil, or exit pupil / entry pupil). Different lens designs result in different pupil ratios, but if you are around a 50 mm focal length, your pupil ratio will be close to 1.0. So it is possible that a longer focal length lens (e.g. 105 mm vs 65 mm) could end up with a pupil ratio that is not equal to 1.0, so the effective aperture for the longer focal length lens could be different even though both the 105 mm lens and the 65 mm lens are set to f/8 (and the magnification is equal).
So "in real life" you could see a difference in DOF between a 65 mm lens and a 105 mm lens, but this difference will be due to the effective aperture difference, not the directly as a result of the focal length difference.
