Such a Sylow $3$-subgroup is a direct product of "iterated wreath products" of the cyclic group of order $3$. It depends on the base $3$ expansion of $n.$ If $n = a_{0} + 3 a_{1} + \ldots + 3^{m-1}a_{m-1}$ where each $a_{i} \in \{0,2 \}$, then a Sylow $3$-subgroup of $S_{n}$ is the direct product over $i$ of a direct product of $a_{i}$ copies the same as a Sylow $3$-subgroup of $S_{3^{i}}$. This reduces us to considering the case $n = 3^{m}$ for some $m$.
In that case, one takes a direct product of $3$ copies of a Sylow $3$-subgroup of $S_{3^{m-1}},$ and then adjoins a product of $3^{m-1}$ $3$-cycles interchanging the sets of points moved by the respective direct factors in a way compatible with their respective actions. This is more complicated to say than to implement: for example, to go from $S_{3}$ to $S_{9},$ take $\langle (123) \rangle \times \langle (456) \rangle \times \langle (789) \rangle$ and adjoin $(147)(258)(369).$
There is nothing in this construction which is particular to the prime $3$. The obvious analogues work for Sylow $p$-subgroups of $S_{n}$ for any prime $p.$