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, but I think it points out some underlying ideas common to both approaches). I do not give all details,and while I make mention of relevant background results from 
time to time, the treatment is essentially self-contained (apart from the use of the simplicity of ${\rm Sp}(2n,p)$).

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 $T = {\rm SU}(p^{n},\mathbb{C})$.

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})$, as we will outline here, summarizing and amalgamating the results and methods of many authors, as referenced,for example, in @spin's answer . 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). For $C_{T}(E\rho)$ has order $p$, and $N_{T}(E\rho)/C_{T}(E\rho)$ is isomorphic to a subgroup of the finite group ${\rm Aut}(E).$  

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 $p$).
This simplifying remark (no pun intended) is not essential, but it makes the exposition easier if we assume it.

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 forced by determinant one,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 also 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 ${\rm Sp}(2n,p)$ we have made use of the fact that ${\rm Sp}(2n,p)$ is generated by its elements of order prime to $p$ (also known as $p$-regular elements)

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 $4$) and $\epsilon = \pm 1$.

 This gives that $N_{T}(E) = EC_{T}(\sigma)^{\prime}$ for this choice of $\sigma$, since we have $N_{T}(E) \cap C_{T}(\sigma) \cong Z(E) \times {\rm Sp}(2n,p)$ ( here,the superscript used denotes taking the derived subgroup, as usual).

The uniqueness of $\sigma\rho$ for a chosen general automorphism $\sigma \in {\rm Sp}(2n,p)$ of order $h$, coprime to $p$, can be seen as follows, which is a (familiar) explicit way to define a matrix with the correct intertwining properties ( unique up to scalars, due to irreducibility of the representation of $E$, as noted in the question).

We define the matrix $M(\sigma)$ via $M(\sigma) = \sum_{u \in E/C_{E}(\sigma)}
(u^{-1}u^{\sigma})\rho$ ( 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 $(v\rho)^{-1}M(\sigma) (v^{\sigma}\rho) = M(\sigma)$ for each $v \in E$, so
that $M(\sigma)^{-1}(v\rho) M(\sigma) = (v^{\sigma} \rho)$ for each $v \in E$, so conjugation by the matrix $M(\sigma)$ has the same effect on $E\rho$ as $\sigma$ does on $E$.

We may note that $M(\sigma)$ really is invertible as follows. The only time that $(u^{-1}u^{\sigma})\rho$ can have non-zero trace is 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 $u^{\sigma} = u$ in that case. It follows then that ${\rm trace}(M(\sigma) ) = p^{n} = \chi(1).$ In particular, $M(\sigma)$ is not nilpotent.

Since $\sigma$ is an automorphism of order $h$ coprime to $p$, it follows that $M(\sigma)^{h}$ is a non-zero scalar matrix. Hence we may take a scalar multiple of $T(\sigma)$ of multiplicative order $h$ (in fact an easy character-theoretic argument tells us that $M(\sigma) = \epsilon(\sigma)p^{n-d(\sigma)} F(\sigma)$, where $F(\sigma)$ is a matrix of 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 $\sigma \rho = F(\sigma).$ This is because we find that ${\rm trace}(M(\sigma)) = [E:C_{E}(\sigma)] \frac{| {\rm trace}(F(\sigma)|^{2}} {\chi(1)},$ while we know that  $M(\sigma)$ has trace
$p^{n}.$

While it might appear that the necessary choice of scalar is not explicit, and outside our control, the scalar is in fact uniquely recoverable from the representation-theoretic and character-theoretic
information we have available. 

The choice of $\sigma \rho$ as $F(\sigma)$ tells us that we have extended the character $\chi$ to $E\langle \sigma\rangle$ in such a way that $M(\sigma) = [E:C_{E}(\sigma)] \frac{\overline{\chi(\sigma)}}{\chi(1)}(\sigma \rho)$, and also that $|\chi(\sigma)|^{2} = \frac{|C_{E}(\sigma)|}{p}.$ In other words, the explicit multiple of $M(\sigma)$ need to obtain a matrix $F(\sigma)$ of finite order $h$ is determined up to a scalar (of absolute value one) multiple, and forcing the determinant to be one fixes the choice of scalar.

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 (given that it has finite order).

Hence the uniquess of the extension of $\rho$ to a unimodular representation of $E\langle \sigma \rangle$ is established whenever $\sigma \in {\rm Sp}(2n,p)$ is $p$-regular, and the demonstration that 
$N_{T}(E\rho)/E\rho$ is isomorphic to ${\rm Sp}(2n,p)$ is complete.

Further edit: I realise that the fact that $p$-regular elements of ${\rm Sp}(2n,p)$ have rational trace in the above representation is not really fully explained. This can be explained at the level of characters, by arguments which I think originate with Isaacs.

At the level of characters, the irreducible character $\chi$ of $E$ has been shown to have a unique extension to an irreducible character of $E{\rm Sp}(2n,p)$ (which we still call $\chi$) such that the well- defined linear character ${\rm det} \chi$ is the trivial character.

Now we note that if $\tau$ is a Galois automorphism (of the appropriate cyclotomic field) which fixes $p$-power roots of unity, then the irreducible character $\chi^{\tau}$ has the same properties as $\chi$, so must be equal to $\chi$ by the uniqueness of $\chi$. Then   we see by Galois theory that $\chi(g) \in \mathbb{Q}$ whenever $g$ is $p$-regular.