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).