How to prove that $||f||_{L^\infty(\mathbb R^n)}\le\frac 1{2^n}\int_\mathbb {R^n}|\frac {\partial f}{\partial x_1\cdots \partial x_n}|$?

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Consider $f\in W^{n,1}(\mathbb R^n)$. I want to show that $||f||_{L^\infty(\mathbb R^n)}\le\frac 1{2^n}\int_\mathbb {R^n}|\frac {\partial f}{\partial x_1\cdots \partial x_n}|$.

I tried to use, maybe, one of the Sobolev inequalities for functions $f \in C^n(\mathbb R^n)$ with compact support and then to use the density of this space in $W^{n,1}(\mathbb R^n)$. Any suggestions?

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Start with $f$ that is smooth with compact support and write

$$f(x) = \int_{-\infty}^{x_1}f_{x_1}(s,x_2,\dots,x_n)\, ds.$$

Then repeat this trick for the integrand $n-1$ more times in each of the $x_2,\dots,x_n$ coordinate directions. Use density of $C^\infty_c(\mathbb{R}^n)$ in $W^{n,1}$ to extend to general $f\in W^{n,1}$.

EDIT: The argument above gives a constant of $1$ instead of $2^n$. An idea to fix this is to note that we also have

$$f(x) = -\int_{x_1}^{\infty}f_{x_1}(s,x_2,\dots,x_n)\, ds.$$

Taking absolute values and adding these identities gives

$$2|f(x)|\leq \int_{-\infty}^\infty |f_{x_1}(s,x_2,\dots,x_n)|\, ds.$$

Then repeat as above and you should get a factor of $1/2^n$.