Surface area of an $\ell_p$ unit ball? Are there any known formulas or approximations for the surface area of a unit ball in $d$ dimensions under the $\ell_p$ norm?  As obvious examples, it is of course well-known that the surface area of a $d$-dimensional unit ball in $\ell_2$ is $\frac{2\pi^{d/2}}{\Gamma(d/2)}$ and that the surface area of a $d$-dimensional unit box (which we could also call the "unit ball under the $\ell_\infty$ norm") is $d\cdot2^d$.
 A: Note: I agree with the criticism of Sawin mentioned by Wong: the information about the volume of the unit ball of $\ell_p^n$ (which you can find, for example, on p. 47 of  Brazitikos, Silouanos; Giannopoulos, Apostolos; Valettas, Petros; Vritsiou, Beatrice-Helen Geometry of isotropic convex bodies. American Mathematical Society, 2014) does not lead (at least immediately) to the answer.   
What I suggest below is not an answer, it is just an information which can be useful in this context; and everyone who rated my posting is expected to reconsider the rating (I also deleted my erroneous comment).
For subsets of the unit sphere of $\ell_p^n$ one can consider the usual surface measure and the cone measure, which is the measure of the corresponding part of the ball. It turns out that if we normalize both measures, the results will be rather close to each other; the best known to me result on this matter is due to Naor (Trans. Amer. Math. Soc. 359 (2007), no. 3, 1045–1079).
A: $\newcommand{\R}{\mathbb R}
\newcommand{\dd}{\operatorname{d}\!}$
This is not a complete answer, but it may lead to finding the asymptotics of the surface area; an exact closed-form expression for the area seems unlikely to exist. (This answer is almost the same as my answer to almost the same question.) It is easy to see that for $p\in(1,\infty)\setminus\{2\}$ this area is given by the formula 
\begin{align*}
 \mathrm{vol}_{d-1}(\partial B_p^d)
 &=2^d\int_{S_p^{d-1}}\frac{\dd x_1\cdots\dd x_{d-1}}{\cos\theta} \\ 
  &=2^d\int_{S_p^{d-1}} \sqrt{1+\Big(1-\sum_1^{d-1}x_i^p\Big)^{2/p-2}\,\sum_1^{d-1}x_i^{2p-2}}\,
  \dd x_1\cdots\dd x_{d-1} \\ 
&=2^d\iint\limits_{a<1} \sqrt{1+(1-a)^{2/p-2}\,b}\,\,f_{d-1}(a,b)\dd a\dd b,  
\end{align*}
where $S_p^{d-1}:=\{(x_1,\dots,x_{d-1})\in(0,\infty)^{d-1}\colon\sum_1^{d-1}x_i^p<1\}$, $\theta$ is the angle between the vectors $(0,\dots,0,1)\in\R^d$ and 
$\vec\nabla G=\big(-\frac{\partial F}{\partial x_1},\dots,-\frac{\partial F}{\partial x_{d-1}},1\big)$, $G:=G(x_1,\dots,x_d):=x_d-F(x_1,\dots,x_{d-1})$, $F:=F(x_1,\dots,x_{d-1}):=\big(1-\sum_1^{d-1}x_i^p\big)^{1/p}$ 
(so that $(0,\infty)^d\cap\partial B_p^d=\{(x_1,\dots,x_d)\in(0,\infty)^d\colon G(x_1,\dots,x_d)=0\}$), $f_{d-1}$ is the joint probability density function (pdf) of the random variables (r.v.'s) $A:=\sum_1^{d-1}X_i^p$ and $B:=\sum_1^{d-1}X_i^{2p-2}$, and $X_1,\dots,X_{d-1}$ are independent r.v.'s uniformly distributed on the interval $(0,1)$. 
Thus, it appears to mainly remain to find an appropriate asymptotics of $f_{d-1}(a,b)$ for $a<1$ and $d\to\infty$. This task may not seem exceedingly difficult, but it has eluded my efforts so far; perhaps someone else will be able to succeed here. 
