Consider the following commutative diagram:

$$\begin{aligned}
&&&& 0\qquad \qquad&\to &0\qquad\qquad && \\
&&&&\downarrow\qquad\qquad && \downarrow\qquad\qquad \\
0 &\to &\Omega_X(\log D)(-D_i)\quad &\to &\Omega_X(\log (D-D_i))\quad &\to &\Omega_{D_i}(\log (D-D_i)|_{D_i})\quad &\to &0\\
&& \downarrow\,\simeq \qquad\qquad &&\downarrow\qquad\qquad && \downarrow\qquad\qquad \\
0 &\to &\Omega_X(\log D)(-D_i)\quad &\to &\Omega_X(\log D)\qquad &\to &\Omega_X(\log D)|_{D_i}\quad &\to &0\\
&&&&\downarrow\qquad\qquad && \downarrow\qquad\qquad \\
&&&& \mathscr O_{D_i}\quad \qquad&\overset\simeq\to & \mathscr O_{D_i}\quad\qquad && \\
\end{aligned}$$

The rows and the middle column are standard short exact sequences for logarithmic differentials. The 9-lemma implies that then the last column is also exact. Then the usual formula for Chern classes of short exact sequences (of locally free sheaves!) shows that
$$
c_{n-1}(\Omega_X(\log D)|_{D_i})=c_{n-1}(\Omega_{D_i}(\log (D-D_i)|_{D_i})).
$$

This looks very close to what you probably need, but it seems to me that there might be a sign difference of $(-1)^{n-1}$. It is possible that I made a mistake somewhere, but in any case, this method should work.