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Every complex representation of $S_n$ is defined over $\mathbb Z$, but, as suggested in the comments, it might be more natural to remember only that they are defined over $\mathbb Q$.

Let $k$ be any characteristic-$0$ field, and consider the representations $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$. Their characters satisfy $$ \sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p [g_j = 1] = [(g_1, \dotsc, g_p) = (1, \dotsc, 1)], $$ using the Iverson bracket, since they do so over $\mathbb C$. In particular, if $\chi$ is any character of an irreducible representation of $S_{n_1} \times \dotsb \times S_{n_p}$ over $k$, then the inner product $\frac1{n_1!\dotsm n_p!}\chi(1)$ with the right-hand side is non-$0$, so the inner product of $\chi$ with some $\chi_1 \otimes \dotsb \otimes \chi_p$ is non-$0$, so $\chi$ equals $\chi_1 \otimes \dotsb \otimes \chi_p$.

(Since we're working over a random characteristic-$0$ field $k$, it may not be obvious what I mean by "inner product". I'll define $\langle f_1, f_2\rangle = \frac1{\lvert G\rvert}\sum_{g \in G} f_1(g^{-1})f_2(g)$. Note, though, that it doesn't really matter that much, since all the functions with which we're dealing are valued in the rational numbers—even the integers.)

I'm inclined to think character theoretically, but here's a re-phrasing in entirely representation-theoretic terms. I'll continue to work over $\mathbb Q$; in particular, I'll regard all representation spaces as at first being over $\mathbb Q$. Then the map \begin{equation} \label{Plancherel} \tag{*} \bigoplus_{\rho_1, \dotsc, \rho_p} (V_1 \otimes \dotsb \otimes V_p)^* \otimes (V_1 \otimes \dotsb \otimes V_p) \to \mathbb Q[S_{n_1} \times \dotsb \times S_{n_p}] \end{equation} sending $v^* \otimes v$ to the matrix coefficient $g \mapsto \langle v^*, g\cdot v\rangle$ becomes an isomorphism when tensored with $\mathbb C$, so is already an isomorphism over $\mathbb Q$, so remains an isomorphism when tensored with $k$. Of course, it is $S_{n_1} \times \dotsb \times S_{n_p}$-equivariant. Now any representation $\rho$ of $G = S_{n_1} \times \dotsb \times S_{n_p}$ over $k$ injects into $k[G]$, by taking matrix coefficients as above, and so, by \eqref{Plancherel}${}\otimes_{\mathbb Q} k$, admits a non-$0$ quotient map to some $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$.

Every complex representation of $S_n$ is defined over $\mathbb Z$, but, as suggested in the comments, it might be more natural to remember only that they are defined over $\mathbb Q$.

Let $k$ be any characteristic-$0$ field, and consider the representations $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$. Their characters $\chi_{k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)} = \chi_1 \otimes \dotsb \chi_p$ satisfy $$ \sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p n_j![g_j = 1] = n_1!\dotsm n_p![(g_1, \dotsc, g_p) = (1, \dotsc, 1)], $$ using the Iverson bracket, since they do so over $\mathbb C$. In particular, if $\chi$ is any character of an irreducible representation of $S_{n_1} \times \dotsb \times S_{n_p}$ over $k$, then the inner product $\chi(1)$ with the right-hand side is non-$0$, so the inner product of $\chi$ with some $\chi_1 \otimes \dotsb \otimes \chi_p$ is non-$0$, so $\chi$ equals $\chi_1 \otimes \dotsb \otimes \chi_p$.

(Since we're working over a random characteristic-$0$ field $k$, it may not be obvious what I mean by "inner product". I'll define $\langle f_1, f_2\rangle = \frac1{\lvert G\rvert}\sum_{g \in G} f_1(g^{-1})f_2(g)$. Note, though, that it doesn't really matter that much, since all the functions with which we're dealing are valued in the rational numbers—even the integers.)

I'm inclined to think character theoretically, but here's a re-phrasing in entirely representation-theoretic terms. I'll continue to work over $\mathbb Q$; in particular, I'll regard all representation spaces as at first being over $\mathbb Q$. Then the map \begin{equation} \label{Plancherel} \tag{*} \bigoplus_{\rho_1, \dotsc, \rho_p} (V_1 \otimes \dotsb \otimes V_p)^* \otimes (V_1 \otimes \dotsb \otimes V_p) \to \mathbb Q[S_{n_1} \times \dotsb \times S_{n_p}] \end{equation} sending $v^* \otimes v$ to the matrix coefficient $g \mapsto \langle v^*, g\cdot v\rangle$ becomes an isomorphism when tensored with $\mathbb C$, so is already an isomorphism over $\mathbb Q$, so remains an isomorphism when tensored with $k$. Of course, it is $S_{n_1} \times \dotsb \times S_{n_p}$-equivariant. Now any representation $\rho$ of $G = S_{n_1} \times \dotsb \times S_{n_p}$ over $k$ injects into $k[G]$, by taking matrix coefficients as above, and so, by \eqref{Plancherel}${}\otimes_{\mathbb Q} k$, admits a non-$0$ quotient map to some $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$.

Every complex representation of $S_n$ is defined over $\mathbb Z$.

Let $k$ be any characteristic-$0$ field, and consider the representations $k \otimes_{\mathbb Z} (\rho_1 \otimes \dotsb \otimes \rho_p)$. Their characters satisfy $$ \sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p [g_j = 1] = [(g_1, \dotsc, g_p) = (1, \dotsc, 1)], $$ using the Iverson bracket, since they do so over $\mathbb C$. In particular, if $\chi$ is any character of an irreducible representation of $S_{n_1} \times \dotsb \times S_{n_p}$ over $k$, then the inner product $\frac1{n_1!\dotsm n_p!}\chi(1)$ with the right-hand side is non-$0$, so the inner product of $\chi$ with some $\chi_1 \otimes \dotsb \otimes \chi_p$ is non-$0$, so $\chi$ equals $\chi_1 \otimes \dotsb \otimes \chi_p$.

(Since we're working over a random characteristic-$0$ field $k$, it may not be obvious what I mean by "inner product". I'll define $\langle f_1, f_2\rangle = \frac1{\lvert G\rvert}\sum_{g \in G} f_1(g^{-1})f_2(g)$. Note, though, that it doesn't really matter that much, since all the functions with which we're dealing are valued in the algebraic integers.)

Every complex representation of $S_n$ is defined over $\mathbb Z$, but, as suggested in the comments, it might be more natural to remember only that they are defined over $\mathbb Q$.

Let $k$ be any characteristic-$0$ field, and consider the representations $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$. Their characters satisfy $$ \sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p [g_j = 1] = [(g_1, \dotsc, g_p) = (1, \dotsc, 1)], $$ using the Iverson bracket, since they do so over $\mathbb C$. In particular, if $\chi$ is any character of an irreducible representation of $S_{n_1} \times \dotsb \times S_{n_p}$ over $k$, then the inner product $\frac1{n_1!\dotsm n_p!}\chi(1)$ with the right-hand side is non-$0$, so the inner product of $\chi$ with some $\chi_1 \otimes \dotsb \otimes \chi_p$ is non-$0$, so $\chi$ equals $\chi_1 \otimes \dotsb \otimes \chi_p$.

(Since we're working over a random characteristic-$0$ field $k$, it may not be obvious what I mean by "inner product". I'll define $\langle f_1, f_2\rangle = \frac1{\lvert G\rvert}\sum_{g \in G} f_1(g^{-1})f_2(g)$. Note, though, that it doesn't really matter that much, since all the functions with which we're dealing are valued in the rational numbers—even the integers.)

I'm inclined to think character theoretically, but here's a re-phrasing in entirely representation-theoretic terms. I'll continue to work over $\mathbb Q$; in particular, I'll regard all representation spaces as at first being over $\mathbb Q$. Then the map \begin{equation} \label{Plancherel} \tag{*} \bigoplus_{\rho_1, \dotsc, \rho_p} (V_1 \otimes \dotsb \otimes V_p)^* \otimes (V_1 \otimes \dotsb \otimes V_p) \to \mathbb Q[S_{n_1} \times \dotsb \times S_{n_p}] \end{equation} sending $v^* \otimes v$ to the matrix coefficient $g \mapsto \langle v^*, g\cdot v\rangle$ becomes an isomorphism when tensored with $\mathbb C$, so is already an isomorphism over $\mathbb Q$, so remains an isomorphism when tensored with $k$. Of course, it is $S_{n_1} \times \dotsb \times S_{n_p}$-equivariant. Now any representation $\rho$ of $G = S_{n_1} \times \dotsb \times S_{n_p}$ over $k$ injects into $k[G]$, by taking matrix coefficients as above, and so, by \eqref{Plancherel}${}\otimes_{\mathbb Q} k$, admits a non-$0$ quotient map to some $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$.

Every complex representation of $S_n$ is defined over $\mathbb Z$.

Let $k$ be any characteristic-$0$ field, and consider the representations $k \otimes_{\mathbb Z} (\rho_1 \otimes \dotsb \otimes \rho_p)$. Their characters satisfy $$ \sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p [g_j = 1] = [(g_1, \dotsc, g_p) = (1, \dotsc, 1)], $$ using the Iverson bracket, since they do so over $\mathbb C$. In particular, if $\chi$ is any character of an irreducible representation of $S_{n_1} \times \dotsb \times S_{n_p}$ over $k$, then the inner product $\chi(1)$ with the right-hand side is positive, so the inner product of $\chi$ with some $\chi_1 \otimes \dotsb \otimes \chi_p$ is positive, so $\chi$ equals $\chi_1 \otimes \dotsb \otimes \chi_p$.

Every complex representation of $S_n$ is defined over $\mathbb Z$.

Let $k$ be any characteristic-$0$ field, and consider the representations $k \otimes_{\mathbb Z} (\rho_1 \otimes \dotsb \otimes \rho_p)$. Their characters satisfy $$ \sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p [g_j = 1] = [(g_1, \dotsc, g_p) = (1, \dotsc, 1)], $$ using the Iverson bracket, since they do so over $\mathbb C$. In particular, if $\chi$ is any character of an irreducible representation of $S_{n_1} \times \dotsb \times S_{n_p}$ over $k$, then the inner product $\frac1{n_1!\dotsm n_p!}\chi(1)$ with the right-hand side is non-$0$, so the inner product of $\chi$ with some $\chi_1 \otimes \dotsb \otimes \chi_p$ is non-$0$, so $\chi$ equals $\chi_1 \otimes \dotsb \otimes \chi_p$.

(Since we're working over a random characteristic-$0$ field $k$, it may not be obvious what I mean by "inner product". I'll define $\langle f_1, f_2\rangle = \frac1{\lvert G\rvert}\sum_{g \in G} f_1(g^{-1})f_2(g)$. Note, though, that it doesn't really matter that much, since all the functions with which we're dealing are valued in the algebraic integers.)