From the answer to this question about the complex roots of $\mathrm{Si}\left(\frac{z}{2}\right)$, we now know that the asymptotic of the $k$-th complex root $z_k = x_k + i y_k $ is:
\begin{align} x_k &\approx 4\pi k - \frac{{\log k}}{{\pi k}} \\ \\ y_k &\approx 2\log x_k + 2\log (\pi /2) \end{align}
The Hadamard product for $\mathrm{Si}\left(\frac{z}{2}\right)$ is:
$$\mathrm{Si}\left(\frac{z}{2}\right) = \frac{z}{2}\,\prod_{z_k} \left(1-\frac{z}{z_k} \right) \tag{1}$$
with $z_k$ taken as quadruples $z_k, -z_k, \overline{z_k}, -\overline{z_k}$.
The logarithmic derivative of the RHS of (1) then becomes:
$$\frac{\mathrm{Si}'}{\mathrm{Si}}\left(\frac{z}{2}\right) = \sum_{z_k} \frac{1}{z-z_k} + \frac{1}{z}\tag{2}$$
hence,
$$\mathrm{Si}'\left(\frac{z}{2}\right) = \frac{\sin\left(\frac{z}{2}\right)}{z} =\mathrm{Si}\left(\frac{z}{2}\right)\,\left( \sum_{z_k} \frac{1}{z-z_k} + \frac{1}{z}\right)\tag{3}$$
So, the regular real roots of $\displaystyle \frac{\sin\left(\frac{z}{2}\right)}{z}$, i.e. $\mu_m = \pm 2\pi m$ with $m \in \mathbb{N}$, will occur when:
$$\sum_{z_k} \frac{1}{z-z_k} = -\frac{1}{z} \tag{4}$$
Question:
Suppose we only know the asymptotic for the complex roots $z_k$ of $\mathrm{Si}\left(\frac{z}{2}\right)$, could we then derive, e.g. using (4), any information/constraints about the location of the real roots $\mu_m$ of its derivative $\mathrm{Si}'\left(\frac{z}{2}\right)$?