$\newcommand{\E}{\mathbb{E}}$
Unfortunately the upper bound $n^{3/2}$ you obtained cannot be improved. In fact 
$$\E \sup_{Q} \langle W, Q Q^T \rangle = \Theta(n^{3/2}).$$

First of all, note that it is enough to bound this quantity for $Z$ a matrix with independent gaussians (as opposed to a symmetric matrix), since $\sqrt{2} W = Z + Z^T + D$ where $Z$ is a matrix with independent gaussian entries, and $D$ is a diagonal matrix with entries $D_{ii} = (\sqrt{2} - 2)Z_{ii}$. It is fairly easy to bound $\mathbb{E} \sup_Q \langle D, Q Q^T \rangle \leq O(\sqrt{n})$, and since for any $Q$ we have $\langle Z^T, Q Q^T\rangle = \langle Z, Q Q^T \rangle$, we get

$$
\E \sup_Q \langle W, Q Q^T \rangle \approx \E \sup_Q \langle Z, Q Q^T \rangle
$$

Moreover note that by restricting our choice to $\theta_i \in \pm \pi/2$ we can get $Q Q^T$ to be any rank one matrix of form $q q^T$ for $q \in \{\pm 1\}^n$. On the other hand, for an upper bound we can use a simple triangle inequality to get
$$
\sup_{q \in \{\pm 1\}^n} \langle Z, q q^T \rangle \leq \sup_{Q} \langle Z, Q Q^T \rangle \leq 2 \sup_{q \in [\pm 1]^n} \langle Z, q q^T \rangle,
$$
i.e. it is enough to understand the maximum inner product of a gaussian matrix, with a rank one matrix of form $q q^T$ for $q \in [\pm 1]^n$. Clearly for any realization $Z$ the supremum is obtained at one of the vertices, so in fact we get
$$
\E \sup_{Q} \langle Z, Q Q^T \rangle \approx \E \sup_{q \in \{\pm 1\}^n} \langle Z, q q^T \rangle.
$$

Let us consider now a gaussian process $X_q := \langle Z, q q^T \rangle$, for $q \in \{\pm 1\}^n$. By the Talagrands majoring measure theorem (see for example section 3.4 in [chaining](https://dash.harvard.edu/bitstream/handle/1/34872844/57723140.pdf?sequence=1&isAllowed=yand) for relevant definitions) we have
$$
\E \sup_{q \in \{\pm 1\}^n} X_q \approx \gamma_2(\{\pm 1\}^n, d_X), \tag{1}
$$
where $d_X : \{\pm 1\}^n \times \{ \pm 1\}^n \to \mathbb{R}_{\geq 0}$ is a pseudo metric given by $d_X(u, v) := \sqrt{\E (X_u - X_v)^2}$. 

Note that for $u, v \in \{\pm 1\}^n$, we have 
$$
\mathbb{E} (X_u - X_v)^2 = \|u u^T - v v^T\|_F^2,
$$
where $\|A\|_F^2 := \sum_{ij} A_{ij}^2$. When $\|u\|_2 = \|v\|_2 = \theta$ an elementary calculation leads to 
$$
\|u u^T - v v^T\|_F \approx \theta \min(\|u - v\|_2, \|u + v\|_2).
$$
This gives on one hand 
$$\gamma_2(\{\pm 1\}^n, d_X) \lesssim \gamma_2(\{\pm 1\}^n, \sqrt{n} d_2) = \sqrt{n} \gamma_2(\{\pm 1\}^n, d_2),\tag{2}$$ where $d_2(u,v) = \|u - v\|_2$, and on the other hand, when restricting ourselves to the set $S = \{1\}^{3n/4} \times \{\pm 1\}^{n/4}$, the metric $d_X(u, v)$ for $u, v \in S$ is proportional to $\theta \|u - v\|_2$, and we get
$$
\gamma_2(\{\pm 1\}^n, d_X) \geq \gamma_2(S, d_X) \approx \sqrt{n} \gamma_2(\{\pm 1\}^{n/4}, d_2) \tag{2'}
$$

Finally, applying the majorizing measures theorem again, we know that $\gamma_2(\{\pm 1\}^n, d_2) \approx \E \sup_{q \in \{\pm 1\}} \langle q, G \rangle$, where $G$ is a gaussian vector in $\mathbb{R}^n$.

This gives 
$$\gamma_2(\{\pm 1\}^n, d_2) \approx \E \|G\|_1 \approx n.\tag{3}$$

Combining (1), (2), (2') and (3) gives
$$
\E \sup_{q \in \{\pm 1\}^n} X_q \approx n^{3/2}.
$$