$$L(p)=\int_0^{2^{-\frac{1}{p}}}(1+(x^{-p}-1)^{1-p})^{\frac{1}{p}}dx, p\in \mathbb{R}, p\ge 1$$
Parametrize $x$ as $\cos^{\frac{2}{p}}(t)$, then we can find a equivalent form of the integral, namely
$$L(p)=\int_{\frac{\pi}{4}}^{{\frac{\pi}{2}}}(\frac{2}{p})\sin(t)\cos(t)(\cos^{2-2p}(t)+\sin(t)^{2-2p})^{\frac{1}{p}}, p\in \mathbb{R}, p\ge 1$$
How to use power series to find the approximate value of this integral function with $p$ as a variable?
I don't know if there's a convenient way to turn this into a power series. The function is rather complicated.
Through the sequence of substitutions I was able to transform the integral to the form:
$$L(p)=\frac{1}{p} \int_0^1 v^{1/p-1}(1+v)^{-1-1/p} (1+v^{p-1})^{1/p} \mathrm{d} v$$
Since $v \leq 1$, each bracket can be expanded into a binomial series, for example:
$$(1+v^{p-1})^{1/p}=\sum_{k=0}^\infty \binom{1/p}{k} v^{(p-1)k}$$
Now we need to find the integral:
$$ \int_0^1 v^{(p-1)k+1/p-1}(1+v)^{-1-1/p} \mathrm{d} v$$
This is a very well known Gauss hypergeometric function (see Wikipedia or other sources).
Finally, we have:
What's the point, you may ask? Turns out, the series converges pretty fast. For example, here's the plot for the original integral (blue) vs the approximation with just $\mathbf{5}$ terms (red):
$$L1(p)=\sum_{k=0}^5 \binom{1/p}{k} \frac{1}{p(p-1)k+1} {_2 F_1} \left((p-1)k+\frac{1}{p},1+\frac{1}{p};(p-1)k+\frac{1}{p}+1;-1 \right)$$