I have (I think) a proof under the assumptions I mentioned in my question, that the nerve is built from a dense subcategory and that there is a corresponding realization functor as a left adjoint. I would very much appreciate any readers of this answer to comment or edit if they find errors in my reasoning; and if you have an idea for a better proof, to post your own answer which I will accept. Thank you!

Let's fix some notation. We will consider a fully faithful, dense inclusion $K : C_K\hookrightarrow C$. This generates a nerve functor $N(c) = C[K\bullet,d] : C \to \widehat{C_K}$; because $K$ is dense, the nerve functor is fully faithful. We assume a left adjoint to $N$, the *realization* $|-|: \widehat{C_K} \to C$. I am only going to use this left adjoint as a crutch at some points to transpose along; it may be that there is a better proof which doesn't use it.

We will now check that a dependent product in $C$ is taken to a dependent product in $\widehat{C_K}$. We fix $f : d\to c,g : e\to d$ and wish to check that $\Pi_fg : C/c$ (if it exists) is taken to $\Pi_{Nf}Ng : \widehat{C_K}/Nc$.

We note that the slice of the presheaf category is in fact another presheaf category, the category of presheaves $\widehat{C_K/Nc}$ on the category of elements $C_K/Nc$. Therefore, we may check the isomorphism by probing against "representables", fixing $x : C_K/Nc$; in this case, $x$ is concretely a map $x : K(\partial_0x)\to c$ in $C$. Let us write $\mathcal{E}/X$ for each slice $\widehat{C_K/X}$.

First, we transpose:
\begin{align*}
\mathcal{E}/Nc[Y(x),N(\Pi_fg)]
&\cong C/c[|Y(x)|, \Pi_fg]
\end{align*}

We note that $|Y(x)| : C/c$ is actually just the map $x$ described above; this is a consequence of the density of $K$.

\begin{align*}
&\cong C/c[x,\Pi_fg]
\\
&\cong C/d[f^*x, g]
\end{align*}

Now, because $N$ is fully faithful, we may just put it everywhere:

\begin{align*}
&\cong \mathcal{E}/Nd[N(f^*x),Ng]
\end{align*}

But $N$ is continuous; so we may commute it into the pullback:

\begin{align*}
&\cong \mathcal{E}/Nd[Nf^*Nx,Ng]
\\
&\cong \mathcal{E}/Nc[Nx,\Pi_{Nf}Ng]
\end{align*}

But $Nx$ is isomorphic to $Yx$ (this follows again from the assumption that $K$ is fully faithful and dense), so we are done.

yes. $\endgroup$ – Tim Campion Nov 3 '19 at 20:19