Okay, after thinking on it more, I think I can prove that $\log 2$ is exact.
Suppose $B\subseteq \Bbb{Z}_2$ is $(n, \varepsilon)$-covering, and WLOG $\varepsilon < \frac12$. Let $A = \{0, 1, 2,\ldots, 2^n-1\}$.
By $(n, \varepsilon)$-covering assumption, for each $r\in A$ there exists a $b \in B$ such that $d_n(r, b) < \varepsilon$. In particular, one can form a function $f:A\to B$ such that $d_n\big(r, f(r)\big) <\varepsilon$ for each $r\in A$.
Claim: $f$ is $1$-$1$. Were it the case, then $f(A) \subseteq B$ implies $2^n = \#A = \#f(A) \leq \#B$
(here $\#S$ denotes the cardinality of a finite set). It implies that $2^n \leq \#B$ for any $(n, \varepsilon)$-covering subset $B$ - in particular, $2^n \leq \operatorname{cov}(n, \varepsilon, T)$, and therefore
\begin{align*}
\log 2 &= \lim_{\varepsilon\to0^+} \ \limsup_{n\to\infty}\ \frac1n \log 2^n \\[0.5em]
&\leq \lim_{\varepsilon\to0^+} \ \limsup_{n\to\infty}\ \frac1n \log \operatorname{cov}(n, \varepsilon, T) \\[0.5em]
&= h(T)
\end{align*}
Proof of the claim: first note that clearly, for any distinct $r, s \in A$ we have $r\not\equiv s \mod 2^n$
So suppose to the contrary there exists a distinct pair $r, s$ such that $f(r) = f(s)$. Then we have
\begin{equation*}
d_n(r, s) \leq d_n\big(r, f(r)\big) + d_n\big(s, f(s)\big) < 2\varepsilon < 1
\end{equation*}
Here we use the assumption that $\varepsilon < \frac12$. Then by the definition of $d_n$,
\begin{equation*}
d\big(T^i(r), T^i(s)\big) < 1 \quad \forall i = 0, 1, 2, \ldots n-1
\end{equation*}
By the definition of the $2$-adic norm, we have
\begin{equation*}
T^i(r)\equiv T^i(s)\mod 2 \quad \forall i = 0, 1, 2, \ldots n-1
\end{equation*}
An aside fact: the (shortened) Collatz map $T$ has the property that parity sequences of length $N$ are unique to residue classes modulo $2^N$.
That is, given any sequence $p_0p_1\ldots p_{N-1}\in\{0,1\}^N$, there exists a unique integer $x \in \{0, 1, 2, \ldots 2^N-1\}$ such that $T^i(x)\ \%\ 2 = p_i$ for each $i = 0,1 , 2, \ldots N-1$ (here $x\ \% \ 2$ is the canonical reduction of $x$ modulo $2$ into $\{0,1\}$)
Put another way,
\begin{gather*}
T^i(x)\equiv T^i(y)\mod 2 \quad \forall i = 0, 1, 2, \ldots n-1\\
\iff\\
x\equiv y \mod 2^n
\end{gather*}
The proof is a simple inductive argument, following from the fact that $\frac32(2k+1) + \frac12 = 0$ and $\frac12(2k) = 0$ each have unique solutions for $k$ modulo $2$. The property extends to all $2$-adic integers, but since $r$ and $s$ are finite integers we can apply this fact without objection.
From this, the conclusion is $r\equiv s \mod 2^n$, which contradicts the fact that $r$ and $s$ are distinct elements of $A$. The lower estimate for $h(T)$ follows by the previous argument, and so therefore $h(T) = \log 2$
Best Answer
Since $f$ is a continuous map on the compact metric space $\mathbb{T}^2$, one form of the variational principle says that $$ h_{top}(f) = \sup_{\mu \in \mathscr{M}^f_e(\mathbb{T}^2)} h_\mu(f) $$ where $\mathscr{M}^f_e(\mathbb{T}^2)$ is the set of all ergodic $f$-invariant probability measures on $\mathbb{T}^2$. So it is enough to show that $h_\mu(f) = 0$ for any ergodic $f$-invariant measure $\mu$.
A key observation is the following. If $L_y = \{(x,y): x \in \mathbb{T}\}$ is the horizontal line in $\mathbb{T}^2$ at height $y$, then each $L_y$ is an $f$-invariant set. Therefore if $\mu$ is any ergodic measure, there is a unique $y$ such that $\mu$ is supported on $L_y$. Now when restricted to the line $L_y$, the action of $f$ is clearly isomorphic to the circle rotation $x \mapsto x+y$ on $\mathbb{T}$. Since $\mu$ is supported on a line, it is naturally identified with a measure on $\mathbb{T}$, and so our system $(\mathbb{T}^2, f, \mu)$ is naturally isomorphic to the circle rotation $(\mathbb{T}, x \mapsto x+y, \mu)$.
It's well known that 1-dimensional circle rotations have zero topological entropy, so in particular any invariant measure for a circle rotation has zero measure entropy as well. Therefore $h_\mu(f) = 0$. Since this holds for any ergodic $f$-invariant measure $\mu$ we are done.