Each term in this sequence will always be a positive integer. Also, if for some $n$, if $a_n$ is odd, then $a_{n+2}= \frac{a_n+7}{2}$, which is less than $a_n$ unless $a_n$ is less than or equal to $7$. So, the smallest element in our sequence must be at most $7$. If we assume for the sake of contradiction that $2,4,5$ or $6$ is the smallest element in our sequence, we'll get a contradiction right away because by manual calculation, iterating the recursion on any of these numbers will take you to $1$ or $3$ eventually. Now we need to decide whether $1, 3$ or $7$ is the smallest element. Note that $2014^{2015^{2016}} \equiv 5 \pmod{7}$, thus all the elements in the sequence will be congruent to either $3$, $5$, or $6$ when taken mod $2$. Therefore, the only element among $\{ 1, 3, 7 \}$ that could ever appear in our list is $\boxed{3}$.