15. Let $f_{1}(x)=\frac{2}{1+x}, f_{n+1}(x)=f_{1}\left(f_{n}(x)\right)$, and $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+2}$. Then $a_{2014}=$
15. $-\left(\frac{1}{2}\right)^{2015}$.

From the problem, we know
$$
\begin{array}{c}
a_{n+1}=\frac{f_{n+1}(0)-1}{f_{n+1}(0)+2}=\frac{f_{1}\left(f_{n}(0)\right)-1}{f_{1}\left(f_{n}(0)\right)+2} \\
=\frac{\frac{2}{1+f_{n}(0)}-1}{\frac{2}{1+f_{n}(0)}+2}=\frac{1-f_{n}(0)}{4+2 f_{n}(0)}=-\frac{1}{2} a_{n} .
\end{array}
$$

Thus, the sequence $\left\{a_{n}\right\}$ is a geometric sequence with the first term $a_{1}=\frac{f_{1}(0)-1}{f_{1}(0)+2}=\frac{1}{4}$ and the common ratio $q=-\frac{1}{2}$.
Therefore, $a_{2014}=a_{1} q^{2013}=\frac{1}{4}\left(-\frac{1}{2}\right)^{2013}=-\left(\frac{1}{2}\right)^{2015}$.