Already the fact that $u$ is a tempered distribution and is weakly harmonic implies that $u$ is a polynomial. Then observe that the only polynomial in $L^p({\mathbb R}^n)$ is zero.
Another way is to note that weakly harmonic functions are smooth (in fact analytic), hence harmonic. In particular, $u$ has pointwise values. If $p=\infty$, Liouville's theorem implies that $u$ is constant, hence identically $0$. Now assume that $1\leq p<\infty$. Let $x\in{\mathbb R}^n$, and let $B(x,r)$ denote the ball of radius $r>0$ centered at $x$. Then by the mean value property we have $$ |u(x)|\leq\frac1{|B(x,r)|}\int_{B(x,r)}|u|\leq \frac1{|B(x,r)|}\|u\|_{L^p(B(x,r))}|B(x,r)|^{1-1/p}\leq C r^{-n/p}, $$ which, upon taking $r\to\infty$, shows that $u(x)=0$. Here $|B|$ denotes the volume of $B$, and we have used Hölder's inequality. Since $x$ was arbitrary, we conclude that $u\equiv0$.