Here, $$M_\delta[g](x) = \sup_{x \in Q} \left(\frac{1}{|Q|}\int_{Q}g(y)^{\delta}dx\right)^{\frac1\delta}.$$ Let $f^* = M_1[f]$, and denote $w(E) := \int_{E}w(y)dy$. By a CZO, I mean that $T$ is a Calderon-Zygmund Operator, and so is in particular bounded from $L^p(wdx)$ to $L^p(wdx)$ for all $1<p<\infty$. By $dy$ and $dx$ I mean integration with respect to the Lebesgue measure.
It is clear that for all $f \in L^1(wdx)\cap L^2(wdx)$, that: $$ w( M_\delta[T[f]] > \lambda) \lesssim \frac{\|f\|_{L^1(wdx)}}{\lambda}.$$ Indeed, if we let $E = \{ M_\delta[T[f]] > \lambda\}$, then $$w(E) < w((Tf)^* > \lambda) \lesssim \frac{ \|T[f]\|_{L^2(wdx)}^2}{\lambda^2} \lesssim \frac{\|f\|_{L^2(wdx)}^2}{\lambda^2} < \infty,$$ so that $E$ has finite measure (with respect to $wdx$) and Komolgorov's inequality is available.
Hence, $w(E) = w([T[f]^\delta]^* > \lambda^\delta) \lesssim \frac{\int_E\|T[f]\|^\delta}{\lambda^\delta} < \lambda^{-\delta}w(E)^{1-\delta}[T[f]]_{L^{1,\infty}(wdx)}^{ \delta} < \lambda^{-\delta}w(E)^{1-\delta}\|f\|_{L^{1}(wdx)}^{ \delta}$ Rearranging yields $w(E) \lesssim \frac{\|f\|_{L^1(wdx)}}{\lambda}$.
I want to improve this inequality so that it holds for all $f \in L^1(wdx)$ not just $f \in L^1(wdx)\cap L^2(wdx)$. When $w = 1$ this can be done by using the boundedness of $T: H^1 \rightarrow L^1(dx)$. For general $w \in A_1(dx) $ It should also be true, but I fail to see why. Presumably a density argument works, but I can't seem to find it.
Here's an argument that might work: Let $f_n=f*\eta_n$, for some nice approximate identity $(\eta_n)_n$.
Since $Tf_n\to Tf$ in measure, we may assume that $Tf_n\to Tf$ a.e. as well (take a subsequence and label it the same as the original). By Fatou's Lemma and definition of $\sup$, we have that $M_\delta(Tf)\leq \liminf_n M_\delta(Tf_n)$ a.e. Finally, recall that $g$ being $L^{1,\infty}$ is equivalent to $g^\delta\in L^{1/\delta,\infty}$ and $$ \Vert g\Vert_{L^{1,\infty}} \approx \Vert g^\delta\Vert^{1/\delta}_{L^{1/\delta,\infty}}. $$
We'll be using a characterization of weak $L^p$ spaces in terms of a 'Hölder inequality for sets', see this question (set $r=1$ and $p=1/\delta>1$)
Fix a set of finite measure $A$, then using Fatou's Lemma: \begin{align} \int_A M([Tf]^\delta) wdx & \leq \int_A \liminf_n M([Tf_n]^\delta) wdx \\ & \leq \liminf_n \int_A M([Tf_n]^\delta) wdx \\ &\leq\liminf_n \Vert M([Tf_n]^\delta)\Vert_{L^{1/\delta,\infty}} w(A)^{1-\delta}\\ & \leq w(A)^{1-\delta}\liminf_n \| M_\delta(Tf_n)\|^\delta_{L^1,\infty}. \end{align} This gives that $\Vert M([Tf]^\delta)\Vert_{L^{1/\delta,\infty}}\leq \liminf_n \Vert M_\delta(Tf_n)\|^\delta_{L^{1,\infty}}$, and from here it's easy to conclude using the bound you have.