= Solution
Let $c$ be the center and $\ell$ the side length of $Q$. The zero integral of $f$ lets us subtract a constant kernel value:
$$
T(f)(x)=\int_Q\bigl[K(x-y)-K(x-c)\bigr]f(y)\,dy.
$$
Choose $L=4\sqrt d$. For $x$ outside the concentric <cube> $\widehat Q$ of side length $L\ell$, and for $y\in Q$,
$$
|x-c|\geq\frac{L\ell}{2}=2\sqrt d\,\ell,
\qquad 2|y-c|\leq\sqrt d\,\ell.
$$
Thus the domain of integration in $x$ is contained in $\{|x-c|>2|y-c|\}$. After the translations $X=x-c$ and $Y=y-c$, the <Hörmander integral kernel condition> gives
$$
\int_{\widehat Q^c}|K(x-y)-K(x-c)|\,dx\leq C.
$$
If $Y=0$ the difference is zero and the same inequality holds. Apply the <triangle inequality> and <Tonelli theorem> to obtain the <cancellation estimate outside a dilated cube>:
$$
\boxed{\int_{\widehat Q^c}|T(f)(x)|\,dx\leq C\int_Q|f(y)|\,dy=C\|f\|_1.}
$$
In particular the constant on the right is the same $C$ as in the hypothesis; only the dilation $L$ depends on the dimension.
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