\newcommand\de\deltaThe answer is no. In fact, d_H(A,C_n(L_n)) can be however large or even infinite for all n.
Indeed, e.g. suppose that X=[-1,a) for some a\in(0,\infty] endowed with the standard metric |\cdot-\cdot|, \ell(x)=u(x)=-x for x\in[-1,0], \ell(x)=u(x)=0 for x\in[0,a). Let (L_n) be any positive sequence converging to 0, and let p_n=2L_n eventually (that is, for all large enough n). Then A\supseteq[0,a) and C_n(L_n)=[-3L_n,-L_n] eventually, so that eventually
d_H(A,C_n(L_n))\ge|(a-L_n)-(-L_n)|=a
if a<\infty, and d_H(A,C_n(L_n))=\infty if a=\infty.
Details: Let A_n(\de):=\{x\in X\colon d(p_n,[\ell(x),u(x)])\le\de\} and A(\de):=\liminf_{n\to\infty}A_n(\de), so that A=\bigcap_{\de>0}A(\de).
For each \de>0, eventually A_n(\de)\supseteq[0,a), so that A(\de)\supseteq[0,a) for all \de>0 and hence A\supseteq[0,a), as was claimed.
Also, C_n(L_n)=A_n(L_n)=\{x\in[-1,0]\colon |p_n-(-x)|\le L_n\}=[-3L_n,-L_n] eventually, as was claimed, too.