Benjamin Steinberg answered the question, but I wanted to unwind his idea into an explicit formula.

Let $V$ be the representation with basis $e_1,\dots, e_n$, where a permutation $\sigma$ acts by sending $e_i$ to $e_{\sigma(i)}$.

Let $W$ be an irreducible representation of $S_n$.

Fix a linear form $l$ on $W$. We can map $W$ to $V^{\otimes n}$ by sending $w \in W$ to

$$f(w) = \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dots\otimes e_{ \sigma(n) } $$

Then we have $$\sigma' (f(w)) = \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot \sigma'( e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dots e_{ \sigma(n) } )$$ $$= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dots \otimes e_{ \sigma'(\sigma(n))}$$

$$= \sum_{ \sigma \in S_n} l ( (\sigma' \circ \sigma)^{-1}( \sigma'(w)))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dots \otimes e_{ \sigma'(\sigma(n))}$$ $$= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (\sigma' (w) ))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dots\otimes e_{ \sigma(n) } = f(\sigma' (w)) $$

using a change of variables $\sigma \mapsto \sigma' \circ \sigma$ in the last line. So $f$ is a homomorphism, and because $W$ is irreducible, and $f$ is nontrivial (as long as $l$ is a nontrivial linear form, since all the terms in the sum give different basis vectors in the tensor product), this map is the inclusion of a subrepresentation, as desired.

One can get embeddings into $V^{\otimes m}$ for higher values of $m$ by just putting repetitions in the sequence of basis vectors being tensored, or for $n-1$ by removing the last term, but one can't go lower than $n-1$, because of the sign representation (except maybe in characteristic $2$, I guess).

Benjamin Steinberg answered the question, but I wanted to unwind his idea into an explicit formula.

Let $V$ be the representation with basis $e_1,\dotsc, e_n$, where a permutation $\sigma$ acts by sending $e_i$ to $e_{\sigma(i)}$.

Let $W$ be an irreducible representation of $S_n$.

Fix a linear form $l$ on $W$. We can map $W$ to $V^{\otimes n}$ by sending $w \in W$ to $$f(w) = \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) }.$$

Then we have \begin{align*} \sigma' (f(w)) & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot \sigma'( e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm e_{ \sigma(n) } ) \\ & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\ & {}= \sum_{ \sigma \in S_n} l ( (\sigma' \circ \sigma)^{-1}( \sigma'(w)))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\ & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (\sigma' (w) ))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) } = f(\sigma' (w)) \end{align*} using the change of variables $\sigma \mapsto \sigma' \circ \sigma$ in the last line. So $f$ is a homomorphism, and because $W$ is irreducible, and $f$ is nontrivial (as long as $l$ is a nontrivial linear form, since all the terms in the sum give different basis vectors in the tensor product), this map is the inclusion of a subrepresentation, as desired.

One can get embeddings into $V^{\otimes m}$ for higher values of $m$ by just putting repetitions in the sequence of basis vectors being tensored, or for $n-1$ by removing the last term, but one can't go lower than $n-1$, because of the sign representation (except maybe in characteristic $2$, I guess).