I now believe what's really going on here isn't so much a question of Reedyness. Rather, we have the following:
Claim:
Let $\mathcal A \subseteq Cat_n$ be a dense subcategory which is idempotent-complete. Then $\Theta_n \subseteq \mathcal A$.
Here $Cat_n$ means weak $(\infty,n)$-categories.
Remarks:
This does have a "Reedy" upshot, since any Reedy category worth its salt (e.g. any elegant Reedy category) will be idempotent complete. And anyway, density is unaffected by passing to the idempotent completion, so asking for idempotent-completeness is a reasonable "normalization" condition.
We deduce that $\Theta_n$ is the unique minimal idempotent-complete dense subcategory of $Cat_n$, or equivalently that $\Theta_n$ is the intersection of all idempotent-complete dense subcategories of $Cat_n$.
Proof of Claim:
For $A \in \mathcal A$ and $\theta \in \Theta$, let $A' \in Psh_{Set}(\Theta_n)$ be an $n$-quasicategory modeling $A$, and suppose that $\theta$ is not a retract of $A'$. Then any map $A' \to \theta$ factors through $\partial \theta$, i.e. $Hom(A', \partial \theta) \to Hom(A', \theta)$ an isomorphism. So, letting $D \theta$ be a fibrant replacement for $\partial \theta$, we have that $Hom(A', D\theta) \to Hom(A', \theta)$ is a split epimorphism.
Moreover, if $\theta$ is not a retract of $A'$ for any $A \in \mathcal A$, then the isomorphism $Hom(A', \partial \theta) \to Hom(A', \theta)$ is natural in $A'$, and so we get a splitting of $Hom(A', D\theta) \to Hom(A', \theta)$ which is natural in $A'$. That is, the map $\nu_\mathcal{A}(D\theta) \to \nu_\mathcal{A}(\theta)$ is a split epimorphism, where $\nu_\mathcal{A} : Cat_n \to Psh_{Spaces}(\mathcal A)$ is the restricted Yoneda embedding. If $\mathcal A \subseteq Cat_n$ is dense, so that $\nu_\mathcal{A}$ is fully faithful, this implies that $D \theta \to \theta$ is a split epimorphism in $Cat_n$.
That this is generally not the case follows from the following observations:
Notation: For $\theta \in \Theta$, let $H(\theta)$ denote the "long" $n$-fold hom-space of $\theta$, and let $H(D\theta)$ denote the "long" $n$-fold hom-space of $D\theta$ (where $n = dim(\theta)$). (So $H(\theta)$ is contractible -- its unique element is the "big" $n$-morphism of $\theta$ -- the one obtained by pasting together all the other $k$-morphisms of $\theta$ for $k \leq n$. $H(\partial \theta)$ is the boundary of this contractible space.)
Let $\hat \times$ denote the pushout-product of maps of spaces.
Lemma: For $\theta = [k \mid \theta_1, \dots, \theta_k] \in \Theta_n$, we have $H(\theta) \simeq \ast$, $H(\partial [1]) = \emptyset$, and $H(\partial \theta) \simeq (H(\partial \theta_1) \to \ast) \hat \times \cdots \hat \times (H(\partial \theta_k) \to \ast)$.
Proof: That $H(\theta) = \ast$ is obvious and so also that $H(\partial [1]) = \emptyset$. We have $H(\theta) \cong H(\theta_1) \times \cdots \times H(\theta_k)$. The result then follows from the colimit decomposition of $\partial \theta$.
Notation: For $N \in \mathbb N$, let $\underline{N}^n \in \Theta_n$ be defined inductively by $\underline{N}^1 = [N]$, and $\underline{N}^{n+1} = [N \mid \underline{N}^n,\dots,\underline{N}^n]$.
Corollary: For $N \geq 3$ and all $n$, $H(\underline{N}^n)$ is a sphere of finite dimension $\geq 1$.
Proof: When $n = 1$, $H(\underline{N}^1) \simeq S^{N-2}$, as can be computed using the homotopy coherent nerve. Then higher $H(\underline{N}^n)$'s are iterated joins of this space with itself (by the Lemma), so form connected spheres as claimed.
Corollary: For $N \geq 3$ and $\theta = \underline{N}^n$, the map $D\theta \to \theta$ is not a split epimorphism in $Cat_n$.
Proof: Suppose there is a section. On the long hom-space, this is a map from a disk $D^m = H(\theta)$ to a sphere $S^{m-1} = H(D\theta)$. But any section must be the identity on all but the long hom-space, and therefore it must fix the boundary of the disk $D^m$. That is, the section actually induces a retraction from the disk $D^m$ to its boundary, which Brouwer showed is impossible.
Returning to the proof of the main claim, we see that for $N \geq 3$, we have that $\underline{N}^n$ must be a retract of an object of $\mathcal A$, and hence by idempotent-completeness, must lie in $\mathcal A$. As every $\theta \in \Theta_n$ is a retract of some $\underline{N}^n$, it follows that $\theta \in \mathcal A$ as claimed.
