The solution of the differential equation $$\tag{1}x^2=\phi^2(x)\cdot\phi'(x)+\phi(x)+x\cdot\phi'(x) \quad \text{with} \quad \phi(0)=1$$ is
$$\phi(x)=\frac{\sqrt[3]{x^3+\sqrt{x^6+6 x^3+1}+1}}{\sqrt[3]{2}}-\frac{\sqrt[3]{2}
x}{\sqrt[3]{x^3+\sqrt{x^6+6 x^3+1}+1}}$$ Expanded as a series around $x=0$
$$\phi(x)=\sum_{n=0}^\infty a_n\,x^n$$ the coefficients form the sequence
$$\left\{1,-1,0,\frac{2}{3},\frac{2}{3},0,-\frac{10}{9},-\frac{14}{9},0,\frac{274}{81
},\frac{418}{81},0,-\frac{3058}{243},-\frac{4862}{243},0,\frac{37886}{729},\frac
{61742}{729},0,-\frac{1502290}{6561}\right\}$$ which are the same as those given by @Will Jagy in comments.
Using the above expansion to $O\left(x^{19}\right)$, it is extremely good for $|x|\leq \frac 12$ and, over that range, the norm
$$\Phi=\int_{-\frac 12}^{\frac 12} \Big[\phi(x)-\text{series}\Big]^2 \,dx=3.96 \times 10^{-9}$$
If we compute the norm
$$\Phi_n=\int_{-\frac 12}^{\frac 12} \Big[\phi(x)-\text{series}_n\Big]^2 \,dx$$
$$\left(
\begin{array}{cc}
n & \log_{10}(\Phi_n) \\
1,2 & -3.00952 \\
3 & -4.00751 \\
4,5 & -4.66992 \\
6 & -5.41389 \\
7,8 & -5.70164 \\
9 & -6.38166 \\
10,11 & -6.50545 \\
12 & -7.15209 \\
13,14 & -7.18720 \\
15 & -7.81232 \\
16,17 & -7.79253 \\
18 & -8.40241 \\
19,20 & -8.34550 \\
21 & -8.94398 \\
22,23 & -8.86038 \\
24 & -9.45000 \\
25,26 & -9.34632 \\
27 & -9.92883 \\
28,29 & -9.80953 \\
30 & -10.3862
\end{array}
\right)$$
there is a very strong impact of $n$ on the result.
Edit
Looking at the absolute value of the non-zero coefficients $a_n$, they grow exponentially (faster than $e^{\frac n 2}$).
If we compute
$$F(x,\phi(x))=x^3-\phi(x)^3-3x\phi(x)+1$$ Wolfram Alpha expansion gives
$$F(x,\phi(x))=-\frac{3058 }{81}x^{12}+O\left(x^{13}\right)$$ while your gives
$$F(x,\phi(x))=-\frac{64 }{63}x^{10}+O\left(x^{11}\right)$$ and I agree with you that for $|x|\leq \frac 12$
$$\frac{3058 }{81}x^{12} \geq \frac{64 }{63}x^{10} \quad \text{if}\quad |x| \geq \sqrt{\frac{288}{10703}}=0.164$$
Addendum
Not considering them as Taylor series, let
$$\phi_1(x)=1-x+\frac{2}{3}x^3+\frac{2}{3}x^4-\frac{10}{9}x^6-\frac{14}{9}x^7+\frac{274}{81}x^9+\frac{418}{81}x^{10}$$
$$\phi_2(x)=1-x+\frac{2}{3}x^3+\frac{2}{3}x^4-\frac{10}{9}x^6-\frac{14}{9}x^7+\frac{274}{81}x^9+\frac{3118}{567}x^{10}$$
$$F_i=x^3-\phi_i(x)^3-3x\phi_i(x)+1$$
$$F_1-F_2=\frac{64 }{63}x^{10}\, P_{20}(x)$$ and $P_{20}(x)$ does not show any real root. Since its leading coefficient is positive $\Big[\frac{9135556}{321489}\Big]$, $F_1 \geq F_2 ~~\forall ~x$ and you are right !.
We could even play with
$$\phi_3(x)=1-x+\frac{2}{3}x^3+\frac{2}{3}x^4-\frac{10}{9}x^6-\frac{14}{9}x^7+\frac{274}{81}x^9+\left(\frac{418}{81}+\epsilon\right)x^{10}$$
$$F_1-F_3=3x^{10}\,Q_{20}(x)\,\epsilon+3x^{20} \,R_{10}(x)\,\epsilon^2+x^{30}\, \epsilon^3$$ The two polynomials have positive leading coefficients and do not show any real root. So, if $\epsilon >0$, which is the case in your problem $(\epsilon=\frac{64}{189})$, $F_1 \geq F_3 ~~\forall ~x$
Best Answer
Well your series should be
$$\sum_{n=0}^{\infty}\frac{(-1)^n(x^{2n+1})}{(2n+1)!}$$
$$ = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\dots$$
simply plug in $x=\frac{6\pi}{7}$ and $x=\frac{20\pi}{7}$. Then compare the difference (if the series were perfectly accurate they would be the same, because $\frac{20\pi}{7}=\frac{6\pi}{7}+2\pi$)
The error is told by Taylor's inequality
$$|R_n(x)|\leq\frac{M|x-a|^{n+1}}{(n+1)!}$$
where $n$ in this case is the number of terms in your series approximation, and $M$ is the maximum value of what you're looking at (so in the case of sine, the maximum value is 1).
edit: $M$ is actually the maximum $|f^{(n+1)}(x)|\leq M$ where $f$ is the function you're approximating. But in this case that is still $1$, because that will be the maximum of any nth-derivative of $\sin(x)$.
edit 2: yeah it's simply a matter of taking more terms. But it doesn't take long before you need a ridiculous amount of terms to approximate it. For example, if we are looking at something as small as $4\pi$, doing 10 terms gives us an error bound of $30913.828$ by the formula above. It takes up to 35 terms just to get to an error $|R_{35}(x)|\leq 0.0287$.