Estimating the limit $x_{n+1} =x_n - x_{n}^{n+1} $ I wonder whether there is a general method for accurately estimating the limit of the sequence:
\begin{equation}
x_{n+1} = x_n - x_{n}^{n+1}, \forall x_1 \in (0,1)
\end{equation}
After showing that the limit exists, since $ x_n $ is decreasing and bounded, I managed to derive a lower-bound. In particular, I used the fact that:
\begin{equation}
\frac{x_{n+1}}{x_n} = 1-x_{n}^n \tag{1}
\end{equation}
Using $(1)$ we obtain:
\begin{equation}
\frac{x_N}{x_{N-1}}...\frac{x_2}{x_1}=\prod_{n=1}^{N} (1-x_{n}^n)=\frac{x_N}{x_1} \tag{2}
\end{equation}
From this we deduce:
\begin{equation}
\begin{split}
\lim_{N \to \infty} x_N & = \lim_{N \to \infty}x_1 \prod_{n=1}^{N} (1-x_{n}^n) \\  & = x_1 (\lim_{N \to \infty} \prod_{n=1}^{N} e^{\ln (1-x_{n}^n)}) \\ & = x_1 (\lim_{N \to \infty} e^{\sum_{n=1}^N\ln (1-x_{n}^n)})
\end{split}
\tag{3}\end{equation}
Using the following facts:
\begin{cases}
\sum_{n=1}^{N} \ln(1-x_{n}^n) \geq \sum_{n=1}^{N} \ln(1-x_{1}^n),\\
x \approx 0 \implies \ln(1+x) \approx x \\
\tag{4}\end{cases}
We may deduce that for $M$ sufficiently large:
\begin{equation}
\sum_{n=1}^{\infty}\ln (1-x_{n}^n) \geq \sum_{n=1}^{M} \ln(1-x_{1}^n)-\sum_{n=M}^\infty x_{1}^n \tag{5}
\end{equation}
And using $(5)$ we have a useful lower-bound. However, I wonder whether there's a more direct integration technique which can give me a good approximation to $(3)$. 

As you noted, $(x_n) $ is convergent as a decreasing positive sequence.
We should have
$$\lim_{n\to\infty}x_n^{n+1}=$$
$$\lim_{n\to\infty}e^{(n+1)\ln (x_n)}=0$$
for $A <0$ ans great enough $n,$
$$\ln (x_n)<\frac {A}{n+1}$$
