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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.

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).

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 mentioned,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).

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;

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 $N_{T}(E) \cap C_{T}(\sigma) \cong Z(E) \times {\rm Sp}(2n,p)$( here,the superscript $\prime$ denotes taking the derived subgroup, as usual.

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).

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 ${\rm Sp}(2n,p)$).

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 ${\rm Aut}(E).$

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.

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).

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.

The choice of $\sigma \rho$ as $F(\sigma)$ tells us that we have extend 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.

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.