= Solution
Let $\gamma_{\alpha\beta}=\delta_{\alpha\beta}$ be the positive Euclidean spatial background metric. Write a tensor-sector <metric perturbation> as $h_{00}=h_{0\alpha}=0$, $h_{\alpha\beta}=-2E_{\alpha\beta}$, with $\gamma^{\alpha\beta}E_{\alpha\beta}=0$ and $\partial^\alpha E_{\alpha\beta}=0$. Spatial indices here are raised with $\gamma$, and $\Box=\partial_t^2-\Delta$.
Substitute the <linearized Levi-Civita connection> into the cover's curvature definition. To first order,
$$
\delta C_{ab}=\frac12\left(\Box h_{ab}+\partial_a\partial_bh-\partial_a\partial^ch_{bc}-\partial_b\partial^ch_{ac}\right),\qquad
\delta C=\Box h-\partial_a\partial_bh^{ab}.
$$
In the transverse-traceless tensor sector $h=0$ and $\partial^ch_{ac}=0$, so $\delta C=0$ and
$$
\delta G_{\alpha\beta}=\tfrac12\Box h_{\alpha\beta}=-\Box E_{\alpha\beta}.
$$
The tensor part of the <Linearized Einstein equations> therefore gives
$$
\boxed{\ddot E^{\alpha\beta}-\Delta E^{\alpha\beta}=8\pi G\,\delta\widehat T^{\alpha\beta}.}
$$
In a calculation strictly confined to the tensor sector, the source on this equation means its tensor projection; it is already trace-free there.
For a generic compact source, \b[trace-free is not the same as transverse-traceless]. For example, the static compactly supported spatial stress $T^{ij}=\partial^i\partial^j\chi-\gamma^{ij}\Delta\chi$, with smooth compactly supported $\chi$, is divergence-free, but its trace-free part has divergence $2\partial^j\Delta\chi/3$, generally nonzero. There is a precise way to reproduce the unprojected trace-free formula used in the question while retaining this distinction. In <Lorenz gauge in linearized gravity>, the <trace-reversed metric perturbation> $\bar h_{ab}=h_{ab}-\eta_{ab}h/2$ obeys
$$
\partial^a\bar h_{ab}=0,\qquad \delta G_{ab}=\tfrac12\Box\bar h_{ab},\qquad
\Box\bar h_{ab}=-16\pi G\,\delta T_{ab}.
$$
Define its spatial trace-free half-strain by
$$
E_{\alpha\beta}=-\frac12\left(\bar h_{\alpha\beta}-\frac13\gamma_{\alpha\beta}\gamma^{\mu\nu}\bar h_{\mu\nu}\right).
$$
Then its wave equation has exactly the displayed trace-free source $\delta\widehat T^{\alpha\beta}=\delta T^{\alpha\beta}-\gamma^{\alpha\beta}\delta T^\mu{}_{\mu,\mathrm{spatial}}/3$. The physical tensor radiation is obtained afterwards with the <transverse-traceless projector>. Without that projection, this $E$ is a Lorenz-gauge trace-free coefficient, not necessarily a transverse tensor mode. At leading far-zone order, the same directional projector is applied to both sides of the quadrupole field formula below.
The flat <contracted Bianchi identity> implies <stress-energy conservation>, $\partial_b\delta T^{ab}=0$. Let
$$
I^{\alpha\beta}(t)=\int\delta T^{00}(t,\mathbf x)x^\alpha x^\beta\,d^3x.
$$
Compact support removes boundary terms. Differentiate and integrate by parts once:
$$
\begin{aligned}
\dot I^{\alpha\beta}
&=-\int\partial_\mu\delta T^{0\mu}\,x^\alpha x^\beta\,d^3x\\
&=\int\left(\delta T^{0\alpha}x^\beta+\delta T^{0\beta}x^\alpha\right)d^3x.
\end{aligned}
$$
A second time derivative, another integration by parts and symmetry of the <stress-energy tensor> give
$$
\ddot I^{\alpha\beta}
=-\int\left(\partial_\mu\delta T^{\alpha\mu}x^\beta+\partial_\mu\delta T^{\beta\mu}x^\alpha\right)d^3x
=2\int\delta T^{\alpha\beta}\,d^3x.
$$
Take the spatial trace-free part. The <mass quadrupole moment> satisfies the <stress-energy quadrupole identity>
$$
Q^{\alpha\beta}=I^{\alpha\beta}-\frac13\gamma^{\alpha\beta}\gamma_{\mu\nu}I^{\mu\nu},\qquad
\boxed{\ddot Q^{\alpha\beta}=2\int\delta\widehat T^{\alpha\beta}\,d^3x.}
$$
Inserting this into the assumed leading <retarded quadrupole field> gives
$$
\boxed{E^{\alpha\beta}(t,\mathbf x)=\frac{G}{r}\ddot Q^{\alpha\beta}(t-r).}
$$
This is the requested spatial half-strain convention; the measured transverse-traceless strain is $H^{\mathrm{TT}}_{\alpha\beta}=2E^{\mathrm{TT}}_{\alpha\beta}$. The formula selects the retarded source field and does not include an independently added homogeneous <gravitational wave>. Dropping the source-dependent part of retardation also needs the <long-wavelength source approximation>, besides $R/r\ll1$.
For a slowly moving weakly self-gravitating source, the <virial theorem> or the dynamical balance $v^2/R\sim GM/R^2$ gives $v^2\sim\epsilon=GM/R$. A varying nonspherical <mass quadrupole moment> has characteristic size $MR^2$ and dynamical time $\tau\sim R/v$, so $\ddot Q\sim MR^2/\tau^2\sim Mv^2$. Consequently the <quadrupole strain scaling of a weak self-gravitating source> is
$$
\boxed{|E^{\alpha\beta}|=O\!\left(\frac{GM}{r}v^2\right)=O\!\left(\epsilon^2\frac{R}{r}\right).}
$$
This is a characteristic scale, not a nonzero lower bound: a stationary or exactly spherical source does not radiate. Slow motion gives $R/\tau\sim\sqrt\epsilon\ll1$, justifying the long-wavelength approximation; for a radiation-zone strain one also takes $r\gg\tau$.
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