Consider the finite dimensional vector space $V=(\mathbb{R}^N,\|\cdot\|_{p})$, equipped with the usual $l^p$ norm, $1<p<\infty$. Consider a linear subspace $U\subset V$ (not necessarily a trivial subspace $\mathbb{R}^n\subset\mathbb{R}^N=\mathbb{R}^n\times \mathbb{R}^{N-n}$!).
Hahn-Banach tells us that there exists a (nonlinear) isometry $H\colon U^*\to V^*$ such that $H(\phi_1)|_U=\phi_1$. Furthermore, since the unit-sphere is smooth, the mapping is unique (geometrically speaking, $\phi_1\in U^*$ determines a hyperplane (the level-1-set) in $U$ with unique touching point $x$ of the $\|\phi_1\|^{-1}$-sphere restricted to $U$, and two norm-preserving extensions of $\phi_1$ would correspond to two supporting hyperplanes at $x$ of the $\|\phi_1\|^{-1}$-sphere, which contradicts smoothness. Btw: do you have a reference for this?)
I wonder if there is an explicit expression of $H$ (note that I consider specifically the $l^p$ norms), or if there are bounds for its derivatives.
I tried: Let $W$ be any algebraic complement of $U$, $V=U\oplus W$.
We have $V^*\simeq U^*\times W^*$ via $\phi\mapsto (\phi_1=\phi|_U,\phi_2=\phi|_W)$ (injectiveness is clear, plus dimensional argument. Note that the inverse is given by $(\phi_1,\phi_2)\mapsto (u+w\mapsto \phi_1(u)+\phi_2(w)$). The uniqueness of the Hahn-Banach extension is now expressed by the fact that for each $\phi_1$ there exists a unique $H_1(\phi_1)\in W^*$ such that $$ \|\phi_1\|_{U^*}-\|(\phi_1,H_1(\phi_1))\|_{V^*}=0 $$ If $p=2$, we could have chosen an orthogonal complement, and the above equation immediately yields that $H_1(\phi)=0$ But if $p\neq 2$ I don't know how to continue. For example, deriving the above doesn't get me anywhere if $\dim(W^*)=\dim(W)>1$.