The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$.
Note that the minimum value of $\sum_{i=1}^n|x_i|^p$ in the sphere $\mathbb{S}^{n-1}\subseteq\mathbb{R}^n$ is achieved when $x_1=\dots=x_n$. This can be deduced by setting $y_i=x_i^2$ and then $\sum y_i^{p/2}$ is convex, so for any $y_1,\dots,y_n\geq0$, if $\sum y_i=1$ then $\sum y_i^{p/2}\geq n\left(\frac{1}{n}\right)^{p/2}$, with equality when $y_1=\dots=y_n$.
This implies that the ball $B$ of radius $n^\frac{p-2}{2p}$, which is tangent to the $L^p$ ball at the points $(\pm\frac{1}{n^{1/p}},\dots,\pm\frac{1}{n^{1/p}})$, contains the $L^p$ ball. So the $n-1$-volume of $\partial B$ is bigger than that of the $L^p$-ball: this works in general for two convex sets one contained in the other, which can be checked using the same proof as in this MSE answer to Surface area of a convex set less than that of its enclosing sphere?.
Now, the area of $\mathbb{S}^{n-1}$ is $\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}$, so the area of $\partial B$ is $A=\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}\left(n^\frac{p-2}{2p}\right)^{n-1}$. So
$$\ln(A)=-\ln\left(\Gamma\left(\frac{n}{2}\right)\right)+n\frac{p-2}{2p}\ln(n)+O(n),$$
which by Stirling's approximation is $$-\frac{n}{2}\ln\left(\frac{n}{2}\right)+n\ln(n)\frac{p-2}{2p}+O(n)=-\frac{n}{2}\ln(n)+n\ln(n)\frac{p-2}{2p}+O(n),$$ which goes to $-\infty$ as $n$ tends to $\infty$, concluding the proof.