$a_1$ can only be $2^{2015}$.
Let $s_n = a_n + b_n$ and $p_n = a_n b_n$ for all $n \ge 1$. The relations become $s_{n+1} = \frac{s_n}{2}$ and $p_{n+1} = \sqrt{p_n}$. Inductively, we find that $s_n = \frac{s_1}{2^{n-1}}$ and $p_n = \sqrt[2^{n-1}]{p_1}$.
Since $a_n$ and $b_n$ are real roots of $x^2 - s_n x + p_n = 0$, we have $s_n^2 - 4p_n \ge 0$ for any $n$. This implies
$$\frac{s_1^2}{2^{2n-2}} - 4 \sqrt[2^{n-1}]{p_1} \ge 0.$$
If $p_1 > 0$, then the left-hand side approaches $0 - 4 = -4 < 0$ when $n$ goes to infinity. This is a contradiction. Thus, we must have $p_1 = 0$. This yields $p_n = 0$ for any $n \ge 1$, and hence $a_n = 0$ or $b_n = 0$.

As $b_{2016} = 1$, we need $a_1 = 0$. Using $s_{2015} = 2s_{2016} = 2$ and $p_{2015} = 0$, we easily deduce $\{a_{2015}, b_{2015}\} = \{0, 2\}$. Similarly, by backward induction, we find that $\{a_n, b_n\} = \{0, 2^{2016-n}\}$ for $1 \le n \le 2016$. In particular, $a_1 = 2^{2015}$ since $a_1 > 0$.
It is possible that $a_1 = 2^{2015}$. An example of the sequences is $a_1 = 2^{2015}$, $b_1 = 0$, and $a_n = 0$, $b_n = 2^{2016-n}$ for $n \ge 2$.