Definition 1. A metric space (normed space) $X$ has the extension property exactly in case, for all finite $A\subseteq X$ and isometric (linear isometric) $f:A\rightarrow X$, there exists an extension to a global isometry (global linear isometry) $\hat{f}:X\rightarrow X$ with $\hat{f}\hspace{-0.25em}\upharpoonright\hspace{-0.25em} A=f$.
One can define the notion for other structures too. The main point is, maps are restricted to structure preserving partial isomorphisms, and to extend these to global isomorphisms.
Question 1. Does any Banach space $\ell^{p}(\omega)$ for some $p\in[1,\infty]$ exhibit the extension property? If so, for which $p$? (Here $\omega=\mathbb{N}$.)
In particular, I mainly wish to know about the cases $p\in\{1,\infty\}$.
Remark 1. An additional requirement on $f$ is that the linear extension of $f$ be well-defined and linear, that is, for all $n\in\omega$ it holds:
$$\sum_{i<n}c_{i}f(a_{i})=0\Longleftrightarrow\sum_{i<n}c_{i}a_{i}=0$$
for all vectors $(a_{i})_{i<n}\subseteq A$ and scalars $(c_{i})_{i<n}$. (The $\Longleftarrow$-direction relates to well-definedness, and the $\Longrightarrow$-direction refers to injectivity.)
Being able to extend an isometry to an isometry is rare. I don't think any nontrivial Banach space has this property. Here is a proof for $\ell^1$ and $\ell^\infty$, the two spaces of main interest to you. In both cases $e_1,e_2,\dots$ is the standard set of $0-1$ vectors, e.g. $e_2=(0,1,0,0,\dots)$.
The case of $\ell^1$
Let $A=\{0,e_1,e_2\}$. Define $f(0)=0$, $f(e_1)=e_1$, $f(e_2)=-e_1$. This is an isometry. Suppose it extends to an isometry of $\ell^1$. Let $y = f((e_1+e_2)/2)$. Since $(e_1+e_2)/2$ is at distance $1$ from every element of $1$, it follows that $$\|y\|=1, \quad \|y-e_1\|=1, \quad \|y+e_1\|=1$$ This is a contradiction, since the only element for which $\|y-e_1\|=1$ and $\|y+e_1\|=1$ is $y=0$.
The case of $\ell^\infty$
Let $A=\{e_1+e_2,e_1-e_2,-e_1-e_2,-e_1+e_2\}$, i.e., the vertices of the square $(\pm 1,\pm1)$. Let $f$ fix the points $e_1+e_2$ and $e_1-e_2$, and exchange the points $-e_1-e_2$ and $-e_1+e_2$. This is an isometry. Suppose it has an extension to $\ell_\infty$. Let $y=f(e_1/2+e_2)$ and $z = f(e_1/2-e_2)$. Then $$ \|y-(e_1+e_2)\|=\|(e_1/2+e_2)-(e_1+e_2)\|= \frac12 \tag{1}$$ $$ \|y-(-e_1-e_2)\|= \|(e_1/2+e_2)-(-e_1+e_2)\| = \frac32 \tag{2} $$ from where it follows that $y = \frac12e_1+\frac12 e_2+\sum_{k\ge 3} c_k e_k$ with $|c_k|\le \frac12$. Similarly, $$ \|z-(e_1-e_2)\|=\|(e_1/2-e_2)-(e_1-e_2)\|= \frac12 \tag{3}$$ $$ \|z-(-e_1+e_2)\|= \|(e_1/2-e_2)-(-e_1-e_2)\| = \frac32 \tag{4} $$ from where it follows that $z = \frac12e_1-\frac12 e_2+\sum_{k\ge 3} d_k e_k$ with $|d_k|\le \frac12$.
Thus, $\|y-z\|=1$. But this is a contradiction because $$\|(e_1/2+e_2) - (e_1/2-e_2)\|=2$$