We consider the statistical model given by the Lebesgue-density $$p([x_1,\ldots,x_n],\theta)=\frac{1}{2^n}\exp\left(-\sum_{i=1}^n|x_i-\theta|\right),~~x:=[x_1,\ldots,x_n]\in\mathbb{R}^n$$ for fixed $n\in\mathbb{N}$, $n\geq 2$, with unknown parameter $\theta\in\mathbb{R}$.
How can I see directly (without Pitman–Koopman–Darmois) that the arithmetic mean $T(x)=\overline{x}_n$ is not a sufficient statistic for $\theta$?
Using the factorisation theorem would essentially amount to showing that a representation of the form $$\sum_{i=1}^n |x_i-\theta|=g(x)+h(\overline{x}_n,\theta)$$ is impossible - it certainly seems that way but I don't see a very clear argument that I'm not just "not smart enough" to choose $g$ and $h$.
Alternatively, I suppose one could directly compare probabilities; e.g. for $n=2$ we have $$P(X_1\leq 0 \mid \overline{X}_2=t)=P(X_2\geq 2 \mid \overline{X}_2=t)$$ and again it strongly looks like this probability depends on $\theta$, but to be sure that would have to be checked by looking at the resulting conditional densities.
Is there a more effective approach?