What is the order of this pole? $$f(z)=\frac 1{\cos(z^4)-1}$$

$z=0$ is a pole of $f$, and I believe that the Laurent series centred at $0$ is $-\frac 2{z^8}-\frac 16+...$, which looks like the pole is of order $8$, but why does Wolfram Alpha claim that the pole is of order $2$?
You are right. And Wolfram seems to agree.

By Taylor expansion $\cos z=1-z^2/2+O(z^4)$ so
$$
\cos(z^4)-1=-\frac{z^8}{2}+O(z^{16})=-\frac{z^8}{2}(1+o(1)).
$$
Hence
$$
f(z)=\frac{1}{\cos(z^4)-1}=\frac{1}{-\frac{z^8}{2}(1+o(1))}=-\frac{2}{z^8(1+o(1))}\sim-\frac{2}{z^8}
$$
So $\lim_0 z^8f(z)=-2\neq 0$. This means that $g(z)=z^8f(z)$ is holomorphic at $0$ (in a neighborhood of $0$) with $g(0)=-2$. So a power series expansion of $g$ will yield a Laurent expansion of $f$ starting by $-\frac{2}{z^8}$.

So indeed, $0$ is a pole of $f$ of order $8$.
The order of the pole is the number of elements in the finite (!) principal part of the Laurent series of your function (or the highest negative power).
You have 4 poles (solve the equation $(\cos \left(z^4\right)-1)$: $\left\{\left\{x\to -\sqrt[4]{2 \pi } \left(\sqrt[4]{n}\right)\right\},\left\{x\to -i \sqrt[4]{2 \pi } \sqrt[4]{n}\right\},\left\{x\to i \sqrt[4]{2 \pi } \sqrt[4]{n}\right\},\left\{x\to \sqrt[4]{2 \pi } \sqrt[4]{n}\right\}\right\}$.
If you expand the initial function into Laurent series you will get (for example for the last root):
$\frac{5 \left(z-\sqrt[4]{2 \pi }\right)}{32 \sqrt[4]{2} \pi ^{9/4}}+\frac{3}{16\ 2^{3/4} \pi ^{7/4} \left(z-\sqrt[4]{2 \pi }\right)}+\frac{1}{16 \left(\sqrt{2} \pi ^{3/2}\right) \left(z-\sqrt[4]{2 \pi }\right)^2}+\left(-\frac{1}{6}-\frac{19}{128 \pi ^2}\right)+O\left(\left(\text{z}-\sqrt[4]{2 \pi }\right)^2\right)$
The order of this pole is 2. One can conclude that you have 4 poles, each having order 2. So I guess Wolfram Alpha is correct :).