TL;DR: The existence of transversal Hadamard is independent of which transversal Pauli operators are chosen as the logical Pauli operators, as long as they satisfy the appropriate commutation relations. In the particular case above, $H^{\otimes 7}$ is a logical Hadamard for both choices of logical Paulis. The reason the attempt to define it for the second set of logical Paulis failed is that it tried to achieve too much. Namely, we only need $\overline{H}\overline{X}\overline{H}^\dagger=\overline{Z}$ to hold on the code subspace.
Stabilizers
For reference, let's recall the usual generators of Steane code's stabilizer group $\mathcal{S}$
$$
\begin{array}{c|c}
M_1&XXXXIII\\
M_2&XXIIXXI\\
M_3&XIXIXIX\\
M_4&ZZZZIII\\
M_5&ZZIIZZI\\
M_6&ZIZIZIZ\\
\end{array}\tag1
$$
see e.g. Table $3.4$ on page $23$ in Daniel Gottesman's PhD thesis. Note that qubit order here is opposite$^1$ to that in Figure $10.6$ on page $456$ in Nielsen & Chuang.
Logical Pauli operators
There are two choices of logical operators in play here
$$
\begin{align}
\overline{X}&=XXXXXXX\\
\overline{Z}&=ZZZZZZZ
\end{align}\tag2
$$
and
$$
\begin{align}
\tilde{X}&=IIIIXXX=M_1\overline{X}\\
\tilde{Z}&=ZZIIIIZ=M_5M_4\overline{Z}.
\end{align}\tag3
$$
We know that $\overline{H}=HHHHHHH$ is a logical Hadamard for $(2)$. The task is to find a logical Hadamard $\tilde{H}$ for $(3)$. A sufficient condition for this is that $\tilde{H}$ satisfies
$$
\tilde{H}\tilde{X}\tilde{H}^\dagger=\tilde{Z}.\tag4
$$
Logical Hadamard
As alluded to in the question, it might seem that $\tilde{H}$ cannot be transversal. For if $\tilde{H}=A_1A_2A_3A_4A_5A_6A_7$ then $A_1IA_1^\dagger=I=cZ$ for some scalar $c$ which is a contradiction. However, the equality $(4)$ is a sufficient condition, not a necessary one. There is a weaker sufficient condition. Namely, $\tilde{H}$ must preserve the code subspace and $(4)$ must hold on the code subspace. We can thus rewrite $(4)$ as
$$
\tilde{H}\tilde{X}\tilde{H}^\dagger\equiv\tilde{Z}\tag{4'}
$$
where $\equiv$ denotes equality of operators on the code subspace. For any stabilizer $M$ we have $M\equiv I$. Finally, we notice that one can simply set $\tilde{H}:=\overline{H}=H^{\otimes 7}$, because then
$$
\begin{align}
\tilde{H}\tilde{X}\tilde{H}^\dagger&=\overline{H}M_1\overline{X}\overline{H}\\
&=M_4\overline{H}\overline{X}\overline{H}\\
&=M_4\overline{Z}\\
&\equiv M_5M_4\overline{Z}\\
&=\tilde{Z}
\end{align}\tag5
$$
and we already know that $H^{\otimes 7}$ preserves the code subspace. Therefore, $H^{\otimes 7}$ is a logical Hadamard for both choices of logical Pauli operators $(2)$ and $(3)$.
Generalization
The proof above is independent of which particular stabilizers appear in $(3)$, so $H^{\otimes 7}$ is a logical Hadamard for all pairs of logical operators defined as $\tilde{X}=M\overline{X}$ and $\tilde{Z}=N\overline{Z}$ with $M,N\in\mathcal{S}$.
A slightly modified proof also works if the logical operators are defined as $\tilde{X}=M\overline{P}$ and $\tilde{Z}=N\overline{Q}$ with $P,Q\in\{X,Y,Z\}$ and $P$ and $Q$ satisfy the appropriate commutation relations. However, in this more general case the form of the logical Hadamard may change since it may need to incorporate additional single-qubit Cliffords to map between physical Pauli operators, e.g. $S^{\otimes 7}$ to map between $X^{\otimes 7}$ and $Y^{\otimes 7}$. However, this does not affect transversality of the resulting operator.
$^1$ The question doesn't specify the convention used, but some generators in Nielsen & Chuang anti-commute with the candidate (weight three) logical operators while all generators in Gottesman commute with them.
