Evaluating limit (iterated sine function)

Hint: Taylor series suggests that $$\sin_n x=x-\frac{n}{3!}x^3 +o(x^3)$$ Try to prove it by induction.

It's almost obvious that the limit is $n/6$. We can write $$x - \sin_{n} x = (x - \sin x) + (\sin x - \sin_{2}x) +\cdots + (\sin_{n-1}x - \sin_{n}x)\tag{1}$$ and upon division by $x^{3}$ each term tends to $1/6$.

Explanation: Since $\sin x \to 0$ as $x \to 0$, it follows (by repeatedly replacing $x$ with $\sin x$) that $\sin_{n} x \to 0$ as $x \to 0$ for all integers $n > 0$. Again we have the fundamental limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{2_1}$$ Replacing $x$ by $\sin x$ repeatedly we get the following set of limit evaluations \begin{align} \lim_{x \to 0}\frac{\sin_{2}x}{\sin x} &= 1\tag{2_2}\\ \lim_{x \to 0}\frac{\sin_{3}x}{\sin_{2} x} &= 1\tag{2_3}\\ \cdots &= \cdots\notag\\ \lim_{x \to 0}\frac{\sin_{n}x}{\sin_{n - 1} x} &= 1\tag{2_n} \end{align} Multiplying all the equations above we get $$\lim_{x \to 0}\frac{\sin_{n}x}{x} = 1,\, n > 0\tag{3}$$ Using L'Hospital's Rule we know that $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \lim_{x \to 0}\frac{1 - \cos^{2} x}{3x^{2}(1 + \cos x)} = \frac{1}{6}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = \frac{1}{6}\tag{4_1}$$ Replacing $x$ with $\sin_{n - 1}x$ (also define notation $\sin_{0}x = x$) in the above equation we get $$\lim_{x \to 0}\frac{\sin_{n - 1}x - \sin_{n}x}{\sin_{n - 1}^{3}x} = \frac{1}{6}\tag{4_2}$$ and hence using $(3)$ we get $$\lim_{x \to 0}\frac{\sin_{n - 1}x - \sin_{n}x}{x^{3}} = \lim_{x \to 0}\frac{\sin_{n - 1}x - \sin_{n}x}{\sin_{n - 1}^{3}x}\cdot\frac{\sin_{n - 1}^{3}x}{x^{3}} = \frac{1}{6}\tag{4_3}$$ for all integers $n > 0$ where by definition $\sin_{0}x = x$. Hence $$\lim_{x \to 0}\frac{x - \sin_{n}x}{x^{3}} = \sum_{k = 1}^{n}\lim_{x \to 0}\frac{\sin_{k - 1}x - \sin_{k}x}{x^{3}} = \sum_{k = 1}^{n}\frac{1}{6} = \frac{n}{6}\tag{5}$$ In routine problems like these it does not make much sense to use advanced tools like Taylor series when the answer is arrived at instantly by using basic theorems of "algebra of limits". Sadly the trend on MSE is to treat Taylor and LHR as the holy grail of limits and applying them on almost any limit problem (however trivial it might be in reality). For a problem worthy of these advanced tools please see this question.

This is not an answer but it is too long for a comment.

The problem being really interesting, I tried to have an extra information : how is approached the limit ?

Doing almost the same as in Paramanand Singh's and Vim's answers and comments, I used one extra term for the Taylor series of $\sin(x)$ and arrive to $$sin_n(x)=x-\frac n 6 x^3 -\big(\frac n{30}-\frac{n^2}{24}\big) x^5+\cdots$$ which means that $$ \frac{x-\sin_n(x)}{x^3}=\frac n 6+\big(\frac n{30}-\frac{n^2}{24}\big) x^2+\cdots$$

The way to tackle it is this: think of your functions as living in the ring $\Bbb R[[x]]$ and the field $\Bbb R((x))$. These are the formal power series ring and its fraction-field, the field of Laurent series with only finitely many terms with negative exponents. In the ring of power series, you may always cut off by ignoring terms of degree greater than or equal to a given $n$. This amounts to working in $\Bbb R[x]/(x^n)$. You can easily show how $x+ax^3$ and $x+bx^3$ compose together, modulo $(x^4)$. This observation will give you your answer.