Connection between entropy and the set of factors of a sequence Let $a = (a_n)_{n=0}^\infty$ be a bounded real-valued sequence. By a factor of $a$ I mean a finite block $w \in \mathbb R^l$ that appears in $a$, that is, there exists $n \geq 0$ such that $a_n a_{n+1} \dots a_{n+l-1} = w_0w_1 \dots w_{n+l-1}$. Let $A_l \subset \mathbb{R}^l$ denote the set of factors of $a$ of length $l \geq 0$.
We can always ensure that $a$ is produced by some topological dynamical system $(X,T)$ in the sense that $a_n = F(T^n x)$ where $x \in X$ and $F$ is a continuous map on $X$ (let $X$ be the orbit closure of $a$ under the shift operation, and let $T$ be the shift, $x = a$ and $F(x) = x_0$). Assume additionally that $(X,T)$ is minimal.
I'm interested in the connection between the behaviour of $A_l$ and the entropy of $(X,T)$. In particular, I would like to know how the assumption that $(X,T)$ is zero-entropy translates to asymptotics on the size of $A_l$.
The situation is quite simple if $a_n$ takes on only finitely many values. Then it is well-known that $(X,T)$ has entropy zero if and only if the growth of $A_l$ is subexponential: $|A_l| = \exp(o(l))$. More generally, the entropy of $X$ is given by $\lim_{l\to \infty} \log |A_l|/l$. 
When $a_n$ is allowed to take on arbitrary values the situation becomes less clear to me. It seems (although I'm worried I'm missing some technical issues along the way) that zero entropy should be equivalent to:


*

*For any $\epsilon > 0$, the number of boxes with sidelength $\epsilon$ in $\mathbb{R}^l$ needed to cover $A^l$ is $\exp(o_{\epsilon}(l))$. 


This post on Terence Tao's blog mentions a similar-looking condition, but (?) with a different metric on $\mathbb{R}^l$:


*For any $\epsilon > 0$, the balls with radius $\epsilon$ in $\mathbb{R}^l$ needed to cover $A^l$ is $\exp(o_{\epsilon}(l))$.  


Since unit cubes and unit balls in $\mathbb{R}^l$ behave rather differently as $l \to \infty$, I don't see an easy argument showing equivalence between the two (the natural idea of converting $\epsilon$-covering with balls into an $\epsilon'$-covering with boxes and vice versa fails). However, I see no reason why the two couldn't be equivalent either: after all, the sets we are covering have some additional structure.
Are 1. and 2. equivalent? If so, which norms besides $\ell^2$ and $\ell^\infty$ can one take on $\mathbb{R}^l$? If not, what makes $\ell^2$ special? What is a good reference for equivalence of 2. with zero entropy?  
 A: Here's an attempt. Let me restrict to functions with values in $[0,1]$ and my entropies are computed with binary log.
If we consider $X \subset [0,1]^{\mathbb{Z}}$ with the compact topology obtained from the product of standard topologies on $[0,1]$, then all metrics are equivalent, so they give the same notion and value of topological entropy. 
The correct topology in this sense is $\ell_{\infty}$: by refining partitions to be in the natural basis, refining by shifting and using compactness, it is easy to see that it's enough to consider topological entropy with respect to finite open covers of the "alphabet" $[0,1]$. This is the same as using $\ell_{\infty}$ balls, i.e. boxes (and covering by boxes and packing disjoint boxes gives the same entropy by the basic theory).
Now the way I understand your question is that you want to consider a closed invariant set $X \subset [0,1]^{\mathbb{Z}}$ in the same natural topology, and just naively apply the formula of entropy but with some $\ell_p$ distance, $p \geq 1$, used to compare distance of partial orbits. Well, why not. Let's use the one you suggest: how many $\epsilon$-balls do we need to cover the partial orbits restricted to their values in $[0,1]^n$.
I claim that:

It is possible that all these "$\ell_p$-entropies" are positive for $1 \leq p < \infty$, even if topological entropy ("$\ell_\infty$-entropy") is zero, for a minimal subsystem of $[0,1]^{\mathbb{Z}}$.

Proof. Pick for each $n \geq 1$ a finite minimal binary subshift over alphabet $\{0,1\}$ (so a single double-periodic orbit) $X_n$ where the number of words of length $f(n) = 2 \cdot 3^{n^2}$ is $2^{f(n)}$, that is, all subwords of that length occur. Clearly $X_n$ has zero entropy. Now, consider the (infinite) product $X$ of the subshifts $X_n$. Computing the usual topological entropy with respect to any $\epsilon > 0$ you see only finitely many of the subshifts $X_n$, so you see at most the sum of their entropies, $0+0+\cdots+0 = 0$, thus $X$ has zero entropy. If you pick the periods of the $X_n$ to be coprime, this system is minimal. We can realize it inside $[0,1]^{\mathbb{Z}}$ by using digits 0 and 2 in ternary expansions to code the bits from different subshifts, so that roughly we uncover one more layer $X_n$ of $X$ each time $\epsilon$ is divided by $3$ (for $\ell_\infty$ distance).
Now, consider the $\ell_p$-entropy of this system, computed for $\epsilon = 1$. Consider a length $m = f(n)$ for some $n$. Let us ignore everything but the contribution of $X_n$ in the partial orbits $[0,1]^{m}$. Intuitively, this won't make the covering task any easier. Concretely, simplify our partial orbits to be in $\{3^{-n}, 2 \cdot 3^{-n}\}^m$, where each coordinate only depends on $X_n$-value. It is easy to see that this does not increase any $\ell_p$-distances, since if the $X_n$-value is different, then distance is at least $3^{-n}$ in that coordinate no matter what the other $X_j$-values are. Since distances were not increased, it's enough to show that this new set is hard to cover.
Now, we have a set of partial words in $\{3^{-n}, 2 \cdot 3^{-n}\}^m$ that we have to cover by $\ell_p$-balls. By the choice of $X_n$, this set of simplified partial orbits is exactly the set $\{3^{-n}, 2 \cdot 3^{-n}\}^m$ of cardinality $2^m$. A radius-$1$ ball in $\ell_p$-norm covers at most a Hamming ball or radius $3^{pn}$ (where Hamming distance is the usual non-normalized one, considering $\{3^{-n}, 2 \cdot 3^{-n}\}$ as a binary alphabet), and to cover $2^m$ binary words with Hamming balls of radius $3^{pn}$ (which have cardinality $2^{3^{pn}}$), for any $n \geq p$ you need at least
$$ 2^m/2^{3^{pn}} \geq 2^m/2^{3^{n^2}} = 2^{m/2} $$
balls, by the choice of $m = f(n)$, so the $\ell_p$-entropy, as it is lower bounded by $\log_2 2^{m/2} / m = 1/2$, is positive. QED.
In fact this system should have infinite $\ell_p$-entropy for all $1 \leq p < \infty$ since I only considered the contribution of $X_n$ to the entropy, and the contributions of the $X_n$ are essentially independent. I didn't do the math so I'll just claim it and leave as an exercise.
Alternatively, it should be easy to modify it slightly so the $X_n$-contributions directly tend to $\infty$ by using larger alphabets and replacing ternary expansions with $s$-adic expansions for a sequence of bases $s$. Anyway, if the above construction is correct (and I understood your questions), this answers your first question and perhaps to some extent obsoletes the others (?).
Some quick observations, if some $\ell_p$-entropy is zero, then $\ell_\infty$-entropy is zero by a ball comparison argument. And as you yourself note, $\ell_p$-entropy can be zero even in interesting cases, for example when $X$ is a subshift over a finite alphabet $A \subset [0,1]$. You can also presumably do inverse limits as above, but make $X_n$ so small with respect to $X_{n-1}$ that things work out, and get systems where $\ell_p$ entropies are zero but which are not e.g. expansive.
