Probably for the next year

Directly calculating the derivatives of arcsine is difficult, so we'll solve this problem indirectly. First, note that the first derivative of arcsine is: $\frac{\mathrm{d}}{\mathrm{d}x} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$ We expand the right-hand side as a binomial series: $\begin{aligned} \frac{1}{\sqrt{1 - x^2}} &= 1 + \left(-\frac{1}{2}\right)\left(-x^2\right) + \left(-\frac{1}{2} \cdot -\frac{3}{2}\right)\frac{1}{2}\left(-x^2\right)^2 + \left(-\frac{1}{2} \cdot -\frac{3}{2} \cdot -\frac{5}{2}\right)\frac{1}{2 \cdot 3}\left(-x^2\right)^3 + \cdots \\ &= 1 + \left(\frac{1}{2}\right) x^2 + \left(\frac{1 \cdot 3}{2 \cdot 4}\right) x^4 + \left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right) x^6 + \left(\frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8}\right) x^8 + \cdots \end{aligned}$ This can now be integrated term-by-term to give a series for arcsine. $\begin{aligned} \arcsin(x) &= x + \left(\frac{1}{2}\right) \frac{x^3}{3} + \left(\frac{1 \cdot 3}{2 \cdot 4}\right) \frac{x^5}{5} + \left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right) \frac{x^7}{7} + \left(\frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8}\right) \frac{x^9}{9} + \cdots \end{aligned}$ This, then, must be the Taylor (actually Maclaurin) series for arcsine. But, since this is the Taylor series, we have the following equality: $\sum_{n=0}^\infty \left.\frac{\mathrm{d}^n\arcsin(x)}{\mathrm{d}x^n}\right|_{x=0} \frac{x^n}{n!} = x + \left(\frac{1}{2}\right) \frac{x^3}{3} + \left(\frac{1 \cdot 3}{2 \cdot 4}\right) \frac{x^5}{5} + \left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right) \frac{x^7}{7} + \left(\frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8}\right) \frac{x^9}{9} + \cdots$ The coefficients on both sides must be equal. In particular, we note that the coefficients of even powers of $x$ on the right-hand side are all zero. Equating the coefficients of $x^{2016}$ gives us the following: $\left.\frac{\mathrm{d}^{2016}\arcsin(x)}{\mathrm{d} x^{2016}}\right|_{x=0} = 2016! \cdot 0 = \boxed{0}$

'Even'th derivatives are zero, 'odd'th derivatives are 1

If the nth derivative of the sine function is a multiple of 4, it is just equal to itself. In this case its easy to note that the 2016th derivative is a multiple of 4. Hence, 0 should be the answer. :)