Here is my tentative answer. First of all, this answer is all based on the conventional QED, where we know the electron's self-energy is like $\Sigma=\Sigma({\not}{p})$. My original question deals with nonconventional QED where $\Sigma$ may also have momentum dependence on $p^\mu$ (which could be contracted with a background field of some kind, for instance, therefore resulting in an observer scalar quantity). Hopefully, my arguments may be extended to that nonconventional case.
Basically, I cannot find a consistent way of doing $\frac{\partial p^\mu}{\partial {\not}{p}}$, and I suspect there isn't one at all because $\frac{\partial}{\partial {\not}{p}}$ may not be a well defined operation but instead just a practical symbolic notation. Therefore, my approach is to avoid using $\frac{\partial}{\partial {\not}{p}}$ and instead use $\frac{\partial}{\partial p^2}$, which seems to be the meaningful operation.
As far as I could track, in the context mentioned above, the only instance where we need $\frac{\partial}{\partial {\not}{p}}$ is when imposing $\frac{\partial \Sigma}{\partial {\not}{p}}\Big|_{{\not}{p}=m}=0$. This requirement shows up when we ask the propagator $\frac{i}{{\not}{p}-m-\Sigma}$ to have residue at $i$ when ${\not}{p}=m$ once we already have the condition $\Sigma({\not}p=m)=0$ that ensures it's pole is located at $p^2=m^2$. The important point here is that the specific value $i$ arises after a series of loose symbolic operations for taking the residue:
$Residue = \displaystyle \lim_{{\not}{p}\rightarrow m} ({\not}p-m) \frac{i}{{\not}{p}-m-\Sigma}$,
but in the limit we would get $\frac{0}{0}$ because $\Sigma({\not}p=m)=0$. Thus, we use l'Hôpital's rule (by taking $\frac{\partial}{\partial {\not}{p}}$ on numerator and denominator) and in the end we see that imposing $\frac{\partial \Sigma}{\partial {\not}{p}}\Big|_{{\not}{p}=m}=0$ indeed fix the residue at $i$. This is motivated by the wish of having the renormalized propagator to look like the free one (where there is no $\Sigma$).
On the other hand, we can do the same thing in a meaningful way writing the propagator as $i({\not}{p}-m-\Sigma)^{-1} = i\frac{{\not}{p}+m+\Sigma}{p^2-(m+\Sigma)^2}$ and look for the residue,
$Residue = \displaystyle \lim_{p^2\rightarrow m^2} (p^2-m^2) i\frac{{\not}{p}+m+\Sigma}{p^2-(m+\Sigma)^2}$,
and again we would find $\frac{0}{0}$ and then use l'Hôpital, but by means of the $\frac{\partial}{\partial p^2}$ derivative instead. We find that imposing $\frac{\partial \Sigma}{\partial p^2}\Big|_{{\not}{p}=m}=0$ fix the residue at $2m i$. The extra $2m$ factor is indeed the same we would find for the free case ($\Sigma=0$).
The value of the residue is different in both cases first because the dimensionality of the derivatives are different. Second, the latter approach have a pole at $p^2=m^2$ and this makes precise sense, and the former symbolic approach mention before has the pole at ${\not}{p}=m$ in a very loose sense.
The first point is: in the end, both conditions are expected to give precisely the same result. Both conditions are always taken on shell (${\not}{p}=m$), therefore, given the correct dimensionality factor $2m$, derivatives with respect to $p^2$ should be equivalent to ones with respect to ${\not}{p}$. (One way of understand that is by looking at the effect of both derivatives on $p^2={\not}{p}^{\,2}$ after imposing ${\not}{p}=m$.)
The second point is: because $\Sigma$ depends on ${\not}{p}$ it is much easier to work with $\frac{\partial}{\partial {\not}{p}}$ instead of $\frac{\partial}{\partial p^2}$, also because to use the latter we need to know $\frac{\partial p^\mu}{\partial p^2}$, which is not immediately obvious. Once we know the symbolic notation is equivalent to the meaningful one, using $\frac{\partial}{\partial {\not}{p}}$ is more practical for the conventional QED, because all we need to derive depends on ${\not}{p}$ or $p^2={\not}{p}^{\,2}$.
In some model based on nonconventional QED, we may need to know the derivative of $p^\mu$ with respect to either ${\not}{p}$ or $p^2$. The former I cannot figure out (and I suspect it may not make sense at all), but I expect that figuring out $\frac{\partial p^\mu}{\partial p^2}$ should be enough. For the latter task I invoke Lorentz covariance for the result and write $\frac{\partial p^\mu}{\partial p^2}=a(p^2) p^\mu$. Multiplying both sides by $p_\mu$ and further differentiating by $p^2$ leads to the conclusion that $a=\frac{1}{2p^2}$ and, therefore, $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$.
In the conventional QED I can check the consistency of this result. There, we know that the structure of $\Sigma$ is $\Sigma({\not}{p})=A(p^2){\not}{p}-B(p^2)m$. If we use this explicity structure when finding the conditions for the residue of the propagator to be $2mi$, by the meaningful approach, we would find
$\left[\frac{1}{2}A+m^2\left(\frac{\partial A}{\partial p^2}-\frac{\partial B}{\partial p^2}\right) \right]_{{\not}{p}=m}=0$,
and this results does not require any derivative of $p^\mu$ with respect to $p^2$. On the other hand, we already know this result should be equivalent to $\frac{\partial \Sigma}{\partial p^2}\Big|_{{\not}{p}=m}=0$ which, using the explicit form of $\Sigma$ does require derivation of $p^\mu$ with respect to $p^2$. Using $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$ indeed gives the expected result, indicating that the result $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$ is consistent so far.
The next step would be to generalize the discussion to consider $\Sigma$ with a more general form than the mentioned above (including, for instance, factors with $\gamma_5$ and $\sigma^{\mu\nu}$ contracted with $p^\mu$ and suitable background fields). I am able to outline some points:
The result $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$ remains the same.
The condition for the residue using the symbolic notation seemly remains the same, $\frac{\partial \Sigma}{\partial {\not}{p}}\Big|_{{\not}{p}=m}=0$, but it may not be useful as long as we don't know $\frac{\partial p^\mu}{\partial {\not}{p}}$.
It seems to be harder to find the analogous condition by the meaningful approach with $\frac{\partial}{\partial p^2}$ because we need to invert a more complicated $\Sigma$ structure, but nevertheless we would know how to actually apply the resulting condition, because we know how to do the derivative.
The way out, and I may be streching too far, I suspect may be a correspondence like $\left[\frac{\partial}{\partial {\not}{p}}(\dots)\right]_{{\not}{p}=m} = 2m \left[\frac{\partial}{\partial p^2}(\dots)\right]_{{\not}{p}=m}$.
{\not}pyields ${\not}p$, which looks prettier than\not{p}rendered as $\not{p}$ $\endgroup$