Although there are $24$ permutations of the roots,
not all permutations induce an automorphism fixing $\mathbb{Q}(\sqrt{2})$.
If so, then the automorphism group of any separable splitting field would simply be $S_n$.
For example, consider the cyclic permutation $(\alpha,-\alpha,\beta)$.
This cannot induce an automorphism $\sigma$ fixing $\mathbb{Q}(\sqrt{2})$.
If that were to be the case, then the following relation
$$
\alpha+(-\alpha)=0
$$
would become
$$
\begin{align*}
\sigma(\alpha)+\sigma(-\alpha) &=\sigma(0) \\
-\alpha+\beta &= 0
\end{align*}
$$
which is false.
This happens when the roots have some "unexpected" symmetry among one another (in this case if $\alpha$ is a root, then $-\alpha$ is a root).
First, your $\alpha_1,\alpha_2$ are calculated incorrectly.$\newcommand{\rat}{\mathbb{Q}}$
The four roots should be
$$
\begin{align*}
\alpha &= \sqrt{1+\sqrt{1+\sqrt{2}}} \\
\beta &= i\sqrt{-1+\sqrt{1+\sqrt{2}}}
\end{align*}
$$
and $-\alpha, -\beta$.
Then the splitting field is $L=\rat(\alpha,\beta)$.
Note that $\alpha\beta=i2^{1/4}$,
so $L=\rat(\alpha,i2^{1/4})$.
Now, $\rat(\alpha)\subset\mathbb{R}$,
so $\rat(\alpha)\ne L$
and $[L:\rat(\alpha)]\ge 2$.
Also, the quadratic polynomial $x^2+\sqrt{2}\in\rat(\alpha)[x]$ vanishes $i{2^{1/4}}$,
so $[L:\rat(\alpha)]=2$.
This shows that $[L:\rat(\sqrt{2})]=2\cdot 4=8$.$\newcommand{\g}{\gamma}\newcommand{\a}{\alpha}$
Let $\g=i2^{1/4}$.
Let $\sigma\in\text{Aut}(L/\rat(\sqrt{2}))$.
Then $\sigma$ is completely determined by its images of $\a$ and $\g$.
The automorphism $\sigma$ must map each element to its conjugate over $\rat(\sqrt{2})$.
The $\rat(\sqrt{2})$-conjugates of $\a$ are $\a, -\a, \g/\a, -\g/\a$,
and $\rat(\sqrt{2})$-conjugates of $\g$ are $\g,-\g$.
The $8$ automorphisms induced by them account for all $8$ elements in $\text{Aut}(L/\rat(\sqrt{2}))$.$\newcommand{\u}{\mu}\newcommand{\t}{\tau}$
Let $\u$ be the automorphism induced by $\a\mapsto \g/\a$ and $\g\mapsto -\g$.
Let $\t$ be the automorphism induced by
$\a\mapsto\a$ and $\g\mapsto -\g$.
We have the following multiplication table:
$$
\begin{array}{c|cccccccc}
& \text{id} & \u & \u^2 & \u^3 & \t & \t\u & \t\u^2 & \t\u^3\\
\hline
\a & \a & \g/\a & -\a & -\g/\a & \a & -\g/\a & -\a & \g/\a \\
\g & \g & -\g & \g & -\g & -\g & \g & -\g & \g
\end{array}
$$
We see that $\u$ has order $4$ and $\t$ has order $2$,
and $\u\t=\t\u^{-1}$.
Thus, $\text{Aut}(L/\rat(\sqrt{2}))\cong D_8$,
the dihedral group of order $8$.
Best Answer
Let $\alpha=\dfrac1{\sqrt 2}(1+i)$. The roots of $P$ are $\alpha$, $\alpha^3$, $\alpha^5$ and $\alpha^7$. So the splitting field over $\Bbb Q$ is $\Bbb Q(\alpha)$.