I will attempt to write up a representation-theoretic argument influenced by papers I have seen (and sometimes written) over the years, making use of automorphisms of extraspecial groups. It may come under the umbrella of proofs which you have seen, and consider more complicated than the one you have posted (this is certainly more algebraic and less geometric than yours).
If we take an extraspecial group $E$ of order $p^{2n+1}$ and exponent $p$, where $p$ is an odd prime (and we take $p >3$ if $n = 1$, that case needing individual treatment, which we don't pursue here), we may proceed as follows.
The group $E$ has an explicit monomial (and, in particular, unitary) complex irreducible representation of degree $p^{n}$ which is induced from a one-dimensional representation of a maximal Abelian (and normal) subgroup$A$ of $E$. The irreducible character $\chi$ afforded vanishes outside $Z(E)$,and $Z(E)$ has order $p$. This induced representation,say $\rho$ gives an embedding of $E$ into ${\rm SU}(p^{n},\mathbb{C})$.
The representation $\rho$ of $E$ in fact extends uniquely to a homomorphism of the semidirect product $E.{\rm SL}(2n,p)$ into $ T = {\rm SU}(p^{n},\mathbb{C})$. Notice that $N_{T}(E\rho)$ is a finite group (from now on, for ease of notation, we will identify $E$ with $E\rho$) if there is no danger of ambiguity).
The group $N_{T}(E)/E$ is isomorphic to a subgroup of ${\rm Sp}(2n,p)$ ( and is in fact all of ${\rm Sp}(2n,p),$ as will be seen).
Since ${\rm Sp}(2n,p)$ is a non-Abelian simple group ( remember the excluded case!),it is generated by its elements of order prime to $p$).
The idea of the construction to follow is that for any automorphism $\sigma$ of order prime to $p$ of $E$, there is a unique way to extend the representation $\rho$ of $E$ to $E \langle \sigma \rangle $ in such a way that $\sigma \rho \in {\rm SU}(p^{n},\mathbb{C})$ ,and in fact in that case, $\sigma \rho$ has trace $\epsilon(\sigma) p^{d(\sigma)}$, where $\epsilon(\sigma)$ is a unique sign,and $|C_{E}(\sigma)| = p^{2d(\sigma)+1}.$
As I have said in comments,this uniqueness of the unimodular extension to $E\langle \sigma \rangle$ (as far as I know), goes back to Marty Isaacs,( see the answer by @spin), and the fact that the character values of the unique extension are as stated was known to Isaacs. The character values can be calculated using Glauberman correspondence (given the unimodularity)
If we accept this uniqueness for the moment, we see that $\langle E\rho, \sigma \rho$ : $\sigma\in {\rm Sp}(2n,p)$ is $p$-regular $\rangle$ must be all of $N_{T}(E\rho)$ , and that $N_{T}(E)/E$ is (isomorphic to) all of ${\rm Sp}(2n,p).$
Notice in particular that in the case that $\sigma$ is the central element of order $2$ in ${\rm Sp}(2n,p)$, which fixes $Z(E)$ and induces elementwise inversion on $E/Z(E)$, we see that $\sigma \rho$ has trace $\epsilon,$ where $p \equiv \epsilon$ (mod $4$). This gives that $N_{T}(E) = EC_{T}(\sigma)^{\prime}$ for this choice of $\sigma$, since we have $N_{T}(E) \cap C_{T}(\sigma) \cong Z(E) \times {\rm Sp}(2n,p)$.
The uniqueness of $\sigma\rho$ for a chosen general automorphism $\sigma \in {\rm Sp}(2n,p)$ of order $h$ prime to $p$ can be seen as follows,which is a (familiar) explicit way to define a matrix with the correct intertwining properties.
We define the matrix $T(\sigma)$ via $T(\sigma) = \sum_{u \in E/C_{E}(\sigma)} (u^{-1}u^{\sigma})\rho$ ( which is also $|C_{E}(\sigma)|^{-1} \sum_{u \in E}(u^{-1}u^{\sigma})\rho.$
The latter expression makes it clear that we have $(v\rho)^{-1}T(\sigma) (v^{\sigma}\rho) = T(\sigma)$ for each $v \in E$, so that $T(\sigma)^{-1}(v\rho) T(\sigma) = (v^{\sigma} \rho)$ for each $v \in E$.
We note that $T(\sigma)$ is invertible as follows. The only time that $(u^{-1}u^{\sigma})\rho$ can have non-zero traceis when $u^{-1}u^{\sigma} \in Z(E)$,in which case ${\rm trace}(u^{\sigma}\rho) = \lambda {\rm trace}(u\rho)$ for some $p$-th root of unity $\lambda$.
Since $\sigma$ has order prime to $p$, this is easily seen to force $\lambda = 1$, so we need $u^{\sigma} = u$. It follows then that ${\rm trace}(T(\sigma) ) = p^{n}.$ In particular,$T(\sigma)$ is not nilpotent.
Since $\sigma$ is an automorphism of order $h$ prime to $p$, it follows that $T(\sigma)^{h}$ is a non-zero scalar matrix. Hence we may take a scalar mutliple of $T(\sigma)$ of multiplicative order $h$ (in fact an easy character-theoretic argument tells us that $T(\sigma) = \epsilon(\sigma)p^{n-d(\sigma)} F(\sigma)$, where $F(\sigma)$ has order $h$ and determinant one, so the unique way to extend the (special) unitary representation $\rho$ of $E$ to $E\rangle \sigma \rangle$ is to set $\sigma \rho = F(\sigma).$ It is a small exercise to see that general theory and the fact that $E\rho$ consists of unitary matrices( and is irreducible) forces $F(\sigma)$ to be unitary.