Why are $\ell^1$ and $\ell^{\infty}$ dual norms in finite dimension? Surprisingly, I'm not able to find a proof of this after searching for a couple of hours. Why is $\ell^{\infty}$ the dual norm of $\ell^1$ and the other way around. It is easy to see why $\ell^2$ is the dual norm of itself, but not the thing I'm looking for.
 A: $\ell^\infty$ is (isomorphic to) the dual space of $\ell^1$. 
This is very standard material in real analysis. See for instance Folland's Real Analysis. 

If you are only interested in $\ell^1(\mathbb{N})$ but not the general $L^p(X)$ case, see for instance Principle of Functional Analysis by Schechter. 
A: Here is at least an intuitive argument for $\mathbb{R}^n$. If $(v_1,...,v_n)$ is a fixed row vector and $|v_i|$ is maximum among $|v_1|,...,|v_n|,$ then $|v_1 x_1 + ... + v_n x_n|$ is maximized among $(x_1,...,x_n) \in \mathbb{R}^n$ with a fixed sum $|x_1| + ... + |x_n| = 1$ at $$(x_1,...,x_n) = (0,...,1,...,0) = e_i$$ (which you can prove rigorously using Lagrange multipliers if necessary); and the operator norm becomes $$\|(v_1,...,v_n)\| = |(v_1,...,v_n) \cdot e_i| = |v_i| = \max\{|v_1|,...,|v_n|\}.$$ 
Conversely, $|v_1 x_1 + ... + v_n x_n|$ is maximized among $(x_1,...,x_n) \in \mathbb{R}^n$ with a fixed maximum $|x_i| = 1$ at some point of the form $(\pm 1,\pm 1,...,\pm 1)$ (choosing all $\pm$ such that the signs align), in which case the operator norm becomes $$\|(v_1,...,v_n)\| = \|(v_1,...,v_n) \cdot (\pm 1,...,\pm 1)\| = |v_1| + ... + |v_n|.$$
