**Disclaimer.** It turns out that as pointed out by user @Jason Gaitonde, the idea I  presented at the end of my question eventually solves my problem with the right choise of $N_1$, namely $N_1 = C \log N$ for sufficiently large positive constant $C$. In this post, I'll fill in the details. I'd be grateful if someone could kindly check the math. Thanks in advance.

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>**Claim.** *Take $N_1 = 2C \log N$, $k = N/N_1$, where $C$ is a sufficiently large positive constant. For large $N$, it holds w.p $1-1/N^{2C-1} =1-o(1)$ that*
$$
Z(\mathcal T) \gtrsim ((1-\theta)C \log N)^{N/(2C\log N)},
$$

*Proof.* First observe that, in the definision of $S_i$, we can further write $z_t = 2 b_t - 1$, where $b_t$ is Bernoulli with parameter $p=1-\theta/2 \in [1/2,1]$. Thus, $S_i = \sum_{t \in G_i} (2b_t - 1) = 2 B_i - N_1$, where $B_i := \sum_{t \in G_i} b_t \sim \mathrm{Bin}(N_1,p)$. By well-known concentration results, 
\begin{eqnarray}
\begin{split}
\mathbb P(B_i \ge (1+t)N_1 p) &\le e^{-\frac{t^2pN_1}{2+t}},\text{ for all }t > 0,\\
\mathbb P(B_i \le (1-t)N_1 p) &\le e^{-\frac{t^2p N_1}{2}},\text{ for all }0 < t < 1.
\end{split}
\end{eqnarray}
We deduce that,
\begin{eqnarray}
\begin{split}
\mathbb P(S_i \ge (2p(1+t) - 1)N_1) &\le e^{-\frac{t^2p N_1}{2+t}},\text{ for all }t > 0,\\
\mathbb P(S_i \le (2p(1-t)-1)N_1) &\le e^{-\frac{t^2p N_1}{2}},\text{ for all }0 < t < 1.
\end{split}
\end{eqnarray}
Thus, for $t \in (0,a)$, we obtain for any $i$, it holds w.p $1-e^{-t^2 p N_1/2}$ that
$$
S_i \ge ((2p-1)-t) N_1 = (a-t)N_1,
$$
where $a := 2p-1 = 1-\theta \in (1/2,1]$ as before.
A union bound then over $i \in [k]$ then gives: w.p $1-\delta(N_1) = 1 - ke^{-t^2 p N_1/2}$ it holds that
\begin{eqnarray}
\frac{Z(\mathcal T)}{(aN_1)^k} \ge \left(1-t/a\right)^k = \left(\left(1-t/a\right)^{a}\right)^{k/a} \ge e^{-tk/a}.
\end{eqnarray}

Now, we want $k$ to be as large as possible, and the RHS of the above to be as large as possible too. We can achieve this by ensuring that

- $\delta(N_1) = e^{-t^2 p N_1/2 + \log k} = e^{-t^2 p N_1/2 + \log N - \log N_1} \to 0$, and

To satisfy the above constraint (perhaps non-optimally!) it suffices to take
\begin{eqnarray}
N_1 \ge C\log N,\,k=N/N_1 = N/(C\log N),
\end{eqnarray}
where $C$ is sufficiently larger and might depend on $N$. Then, taking $t \in \min(a,(0,\sqrt{2/C}/p)))$, we have $\delta(N_1) = e^{-(t^2 p C/2) \log N + \log N - \log\log N - \log C} = 1/N^{t^2 p C/2-1}$, and so w.p $1-\delta(N_1)$, it holds that
$$
Z(\mathcal T)/(aN_1)^k \gtrsim  b(t)^k
$$
where $b(t) := e^{-t/a} \in (0,1/e)$. In particular, taking $C = 1/(a^2 p^2)$, $N_1 = C \log N$, and $t = a/2$, we deduce that
\begin{eqnarray}
    \begin{split}
Z(\mathcal T) &\gtrsim (a N_1)^k e^{-k/2} = (aCe^{-1/2} \log N)^{N/(C\log N)} \gtrsim 2^{cN\log^2 N / \log N},
  \end{split}
\end{eqnarray}
w.p $1-o(1)$, which proves the claim. $\quad\quad \Box$