This is true (1). It was extended to finitely generated profinite groups here (2). Surprisingly, it is also true in the category of finitely generated modules over a Noetherian commutative ring (3).
(1) Joseph Ayoub, The direct extension theorem, J. Group Theory 9 (2006) 307-316.
(2) Goldstein, Daniel, Guralnick, Robert The direct product theorem for profinite groups. J. Group Theory 9 (2006), no. 3, 317-322.
(3) Takehiko Miyata Note on direct summands of modules. J. Math. Kyoto Univ.
Volume 7, Number 1 (1967), 65-69.
$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\bZ}{\mathbb{Z}}\DeclareMathOperator{\Tor}{Tor}\newcommand{\Tors}{\mathrm{Tors}}$I tried to write up the computation with some level of details, please let me know if anything looks dubious.
Consider the long exact sequence $$\ldots\to H_1(\Gamma, M_2)\to H_1(\Gamma,M_3)\to M_{1,\Gamma}\to M_{2,\Gamma}\to M_{3,\Gamma}\to 0$$ which we will view as an (acyclic) complex and denote its terms by $C^n, n\leq 0$, for brevity, so that $C^0=M_{3,\Gamma},C^{-1}=M_{2,\Gamma}$ etc. Consider the spectral sequence associated with the derived functor of $-\otimes_{\bZ}\bQ/\bZ$ and the 'bête' filtration on the complex $\ldots C^1\to C^0$. I'll use the cohomological grading conventions so that the first page of the spectral sequence looks like $E_{1}^{i,j}=\Tor_{-j}^{\bZ}(C^{-i},\bQ/\bZ)$. Note that for an abelian group $M$ we have $\Tor^{\bZ}_0(M,\bQ/\bZ)=M\otimes\bQ/\bZ,\Tor^{\bZ}_1(M,\bQ/\bZ)=M_{\mathrm{Tors}}$, and $\Tor_{>1}(M,\bQ/\bZ)=0$, hence the $1$st page looks like this, having only $2$ potentially non-zero rows:
$$\begin{matrix}\dots & \color{red}{H_1(\Gamma,M_2)\otimes\bQ/\bZ} & \color{red}{H_1(\Gamma,M_3)\otimes\bQ/\bZ} & M_{1,\Gamma}\otimes\bQ/\bZ & M_{2,\Gamma}\otimes\bQ/\bZ & M_{3,\Gamma}\otimes\bQ/\bZ & \\ \dots & H_1(\Gamma,M_2)_{\mathrm{Tors}} & H_1(\Gamma,M_3)_{\Tors} & M_{1,\Gamma,\Tors} & M_{2,\Gamma,\Tors} & M_{3,\Gamma,\Tors}\end{matrix}$$
The differentials $d_{i,j}:E_1^{i,j}\to E^{i+1,j}_1$ are simply the maps induced by the maps in the above exact sequence. Since we started with an acyclic complex, the spectral sequence must converge to zero.
Now, since $H_i(\Gamma, M)$ is annihilated by $|\Gamma|$ for all $\Gamma$-modules $M$, the two entries that are highlighted in red are in fact zero (and all the hidden entries in the top row are zero likewise). This implies that the only possibly non-trivial differential on the second page is $$E_{2}^{-2,0}=\ker (E_1^{-2,0}\to E_1^{-1,0})=\ker (M_{1,\Gamma}\otimes\bQ/\bZ \xrightarrow{i_*} M_{2,\Gamma}\otimes\bQ/\bZ )\to E_2^{0,-1}=\mathrm{coker} (M_{2,\Gamma,\Tors} \xrightarrow{j_*} M_{3,\Gamma,\Tors})$$
This differential must be an isomorphism, while all other $E_2^{i,j}$ must already be zero as otherwise something would survive to the abutment $E_3^{i,j}=E_{\infty}^{i,j}$ of the spectral sequence. But this is exactly saying that the $6$-term sequence in question is exact, with $\delta$ being defined as the inverse to the differential $E_2^{-2,0}\to E_2^{0,-1}$.
Best Answer
Think of your exact sequence as a resolution of $M_m$. It's not necessarily a resolution by free $G$-modules, or by projective $G$-modules; it's just a resolution by $G$-modules. You get a spectral sequence whose $E_1$-term is the direct sum of the $H_*(G; M_i)$ for all $i<m$, and which converges to $H_*(G; M_m)$.
When the $G$-modules $M_i$ are projective for all $i<m$, then the $E_1$-page degenerates on to a single line, and then running the spectral sequence reduces to the usual process for calculating $Tor$: take a projective resolution, apply the tensor product (at this point you have recovered the one nonzero line in the $E_1$-page), then take homology (at this point you have both the $E_2$-page and the final answer).
This "resolution spectral sequence" is a standard homological tool: see application 5.9.8 in Weibel's homological algebra textbook for its construction.
Having a filtration on the coefficient module $N$ yields a filtration on the bar complex, when you tensor $N$ with the bar resolution of $\mathbb{Z}$ by $G$-modules. The resulting filtered chain complex yields a spectral sequence whose input is homology of $G$ with coefficients in the associated graded $G$-module of your filtration of $N$, and whose output is the homology of $G$ with coefficients in $N$. This was suggested in the comments as well. The resulting spectral sequence is also a standard tool, and although I don't know a textbook that discusses it in exactly this level of generality, any treatment of the spectral sequence of a filtered chain complex will be applicable to this one.