The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$.
Consider the orthoplex circumscribed to an $L^p$-ball, with boundary contained in the $2^n$ hyperplanes $\{\varepsilon_1x_1+\dots+\varepsilon_nx_n=\frac{n}{n^{1/p}}\}$, for $\varepsilon_i\in\{0,1\}$. The $2^n$ points of tangency are $(\pm\frac{1}{n^{1/p}},\dots,\pm\frac{1}{n^{1/p}})$, and the $2n$ vertices of the orthoplex are the points $\pm n^\frac{p-1}{p}$ of all coordinates axes.
Then the $n-1$-volume of the surface of the orthoplex 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?.
So we just have to check that for fixed $p$, the surface area of this orthoplex goes to $0$ when $n$ goes to $\infty$. Note that the orthoplex has edges of length $\sqrt{2}n^\frac{p-1}{p}$, and the surface of the orthoplex is formed by $2^n$ simplices of dimension $n-1$. The volume of an $n-1$-dimensional simplex with unit edges is $\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}$, so the surface volume of the orthoplex is
$$A=2^n\cdot\left(\sqrt{2}n^\frac{p-1}{p}\right)^{n-1}\cdot\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}=2^nn^\frac{(n-1)(p-1)}{p}\frac{\sqrt{n-1}}{(n-1)!}$$
This converges to $0$, as can be seen by taking logarithms:
$$\ln(A)=n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)+\frac{\ln(n-1)}{2}-\ln((n-1)!),$$
which by Stirling's approximation is
$$n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)-(n-1)\ln(n-1)+(n-1)+O(ln(n))$$ $$=\frac{p-1}{p}n\ln(n)-(n-1)\ln(n-1)+O(n)$$ $$=\frac{p-1}{p}(n\ln(n)-(n-1)\ln(n-1))-\frac{1}{p}(n-1)\ln(n-1)+O(n).$$
As $p$ is constant, the term $\frac{1}{p}(n-1)\ln(n-1)$ dominates the others, so the expression goes to $-\infty$ when $n\to\infty$, as we wanted.