Repetitive integration by parts:
\begin{align}
\Phi(-1,1,n+1) &= \int_0^1 \frac{x^n}{x+1} \, {\rm d}x \\
&= \int_0^1 x^{n-1} \frac{x}{x+1} \, {\rm d}x \\
&= \frac{x^n}{n} \, \frac{x}{x+1} \Bigg|_0^1 - \frac{1}{n} \int_0^1 x^{n-1} \left( x \frac{{\rm d}}{{\rm d}x} \right) \frac{x}{x+1} \, {\rm d}x \\
&= \frac{x^n}{n} \, \frac{x}{x+1} \Bigg|_0^1 - \frac{x^{n}}{n^2} \left( x \frac{{\rm d}}{{\rm d}x} \right) \frac{x}{x+1} \Bigg|_0^1 + \frac{1}{n^2} \int_0^1 x^{n-1} \left( x \frac{{\rm d}}{{\rm d}x} \right)^2 \frac{x}{x+1} \, {\rm d}x \\
&=\cdots \\
&=\sum_{k=0}^{N-1} (-1)^k \frac{x^n}{n^{k+1}} \left( x \frac{{\rm d}}{{\rm d}x} \right)^k \frac{x}{x+1} \Bigg|_0^1 + \frac{(-1)^N}{n^N} \int_0^1 x^{n-1} \left( x \frac{{\rm d}}{{\rm d}x} \right)^N \frac{x}{x+1} \, {\rm d}x \, .
\end{align}
As far as I'm aware, the last term is problematic. By the identity
$$\left( x \frac{{\rm d}}{{\rm d}x} \right)^N \frac{x}{x+1} = \sum_{k=1}^N {N\brace k}k! \, x^k \left(\frac{-1}{x+1}\right)^{k+1}$$
with Stirling numbers of the second kind $\left\{\cdot\right\}$, the last integral can be bounded $$\left|\int_0^1 x^{n-1} \left( x \frac{{\rm d}}{{\rm d}x} \right)^N \frac{x}{x+1} \, {\rm d}x\right| \leq \sum_{k=1}^N {N \brace k} (k-1)! \left(1-2^{-k}\right) \leq \sum_{k=0}^N {N\brace k}k! = a(N)$$
where $a(N)$ is the ordered Bell number. Its asymptotics $$a(N) \sim \frac{N!}{2(\log 2)^{N+1}}$$ show that the series is only asymptotic.
Hence, $$\bar{H}_n = \log 2 + (-1)^{n+1} \int_0^1 \frac{x^n}{x+1} \, {\rm d}x \\
=\log 2 + \frac{(-1)^{n+1}}{2n} + (-1)^n \sum_{k=1}^{N-1} \frac{(-1)^k}{n^{k+1}} \sum_{l=1}^k {k\brace l} l! (-1)^{l} 2^{-l-1} + {\cal O}(n^{-N-1})$$
for fixed $N$ and $n\rightarrow \infty$.
E.g. for $N=16$ this becomes $$\bar{H}_n = \log 2 + (-1)^{n+1} \left( \frac{1}{2n} - \frac{1}{4n^2} + \frac{1}{8n^4} - \frac{1}{4n^6} + {\frac {17}{16\,{n}^{8}}}-{\frac {31}{4\,{n}^{10}}}+{\frac {691}{8\,{n}^{12}}}-{
\frac {5461}{4\,{n}^{14}}}+{\frac {929569}{32\,{n}^{16}}}
\right) \, .$$
Another way (which is however similar to the link you gave) I came up with is $(x=e^{-u/n})$:
\begin{align}
\int_0^1 \frac{x^n}{x+1} \, {\rm d}x &= \frac{1}{n} \int_0^\infty \frac{e^{-u}}{e^{u/n}+1} \, {\rm d}u \\
&=\frac{1}{n} \int_0^\infty {\rm d}u \, e^{-u} \sum_{k=0}^\infty (-1)^k e^{-u(k+1)/n} \\
&=\frac{1}{n} \int_0^\infty {\rm d}u \, e^{-u} \sum_{k=0}^\infty (-1)^k \sum_{m=0}^\infty \frac{\left(-u(k+1)/n\right)^m}{m!} \\
&=\frac{1}{n} \int_0^\infty {\rm d}u \, e^{-u} \sum_{m=0}^\infty \frac{\left(-u/n\right)^m}{m!} \, \eta(-m) \\
&=\sum_{m=0}^\infty \frac{\left(-1\right)^m \, \eta(-m)}{n^{m+1}}
\end{align}
where $\eta$ is the Dirichlet $\eta$-function, evaluated by analytic continuation which destroys convergence and makes it only an asymptotic series.
A third method I managed was by contour integration. The basic principle is to express the denominator of the above integrand as
$$\frac{1}{n} \, \frac{1}{e^{u/n}+1} = \frac{1}{2\pi i} \oint_C \frac{1}{e^{s}+1} \, \frac{{\rm d}s}{ns-u}$$
where $C$ is the contour encircling the positive $x$-axis including $0$ counter-clockwise, while leaving out the poles of $\frac{1}{e^s+1}$. The evaluated $u$-integral gives $-e^{-ns} {\rm Ei_1}(-ns)$ and it remains to calculate $$\frac{-1}{2\pi i} \oint_C \frac{{\rm Ei_1}(-ns)}{e^s+1} \, e^{-ns} \, {\rm d}s \, .$$
The trick now is to use ${\rm Ei_1}(z) = {\rm Ein}(z) - \ln(z) - \gamma$, that is since ${\rm Ein}$ is holomorphic the only contribution comes from the logarithm at the cut. Therefore, repeatedly integrating by parts, the last expression becomes
\begin{align}
\frac{1}{2\pi i}\oint_C \frac{\ln(-s)}{e^s+1} \, e^{-ns} \, {\rm d}s &= \frac{1}{2\pi i \, n}\oint_C e^{-ns} \left(\frac{1}{s} + \ln(-s) \frac{{\rm d}}{{\rm d}s} \right) \frac{1}{e^s+1} {\rm d}s \\
&=\frac{1}{2\pi i \, n}\oint_C e^{-ns} \left(\frac{1}{s} + \frac{1}{ns} \frac{{\rm d}}{{\rm d}s} + \frac{\ln(-s)}{n} \frac{{\rm d}^2}{{\rm d}s^2} \right) \frac{1}{e^s+1} {\rm d}s \\
&= \dots \\
&= \frac{1}{2\pi i \, n}\oint_C e^{-ns} \frac{{\rm d}s}{s} \sum_{k=0}^\infty \frac{1}{n^k} \frac{{\rm d}^k}{{\rm d}s^k} \frac{1}{e^s+1} \\
&= \sum_{k=0}^\infty \frac{1}{n^{k+1}} \frac{{\rm d}^k}{{\rm d}s^k} \frac{1}{e^s+1} \Bigg|_{s=0} \, .
\end{align}
Of course the same result would have been immediately obtained by the Borel sum of the Borel transform $\frac{1}{e^s+1}$;
$$\int_0^\infty \frac{e^{-u}}{e^{u/n}+1} \, {\rm d}u = \int_0^\infty {\rm d}u \, e^{-u} \sum_{k=0}^\infty \frac{u^k}{n^k} \, \frac{\frac{{\rm d}^k}{{\rm d}s^k} \frac{1}{e^s+1} \Big|_{s=0}}{k!} = \sum_{k=0}^\infty \frac{1}{n^k} \frac{{\rm d}^k}{{\rm d}s^k} \frac{1}{e^s+1} \Bigg|_{s=0} \, .$$
