Okay, so I mostly worked this out, and I even created lattice diagrams as shown below. But I have a specific question about finding intermediate fields, which I will ask shortly.
Let $\alpha = \sqrt[4]{2}$ and $\omega = e^{\frac{\pi}{4}i} = i$. Then $L = \mathbb{Q}(\alpha, i)$ is the splitting field of $x^4 -2$ over $\mathbb{Q}$. Also, the Galois group $\Gamma_\mathbb{Q}(x^4 - 2) = D_8$ acts on the roots $\alpha, \alpha i, -\alpha,$ and $-\alpha i$, and is generated by rotation $\sigma$ and reflection $\tau$, where $\sigma(i) = i, \sigma(\alpha) = \alpha i$ and $\tau(\alpha) = \alpha, \tau(i) = -i$.
To find the intermediate fields between $L$ and $\mathbb{Q}$, find the subgroups of $D_8$ instead with the idea that finding subgroups is easier and better understood than finding intermediate fields. Then from the subgroups, use the Galois correspondence to get all the intermediate fields.
There are 10 subgroups of $D_8$ which must correspond to 10 intermediate fields. Well, I pieced together 8 obvious candidates for intermediate fields, and in the end, I had to look up the other 2 which were $\mathbb{Q}(\alpha(1 + i))$ and $\mathbb{Q}(\alpha(1 - i))$. Those two seemed strange until I realized that $\sqrt{8\alpha^2 i} = \alpha(1 + i)$.
Finally, I was able to check fixed fields to verify the exact correspondence, and come up with the diagrams.
Question: Is there a systematic approach to finding and connecting up the corresponding intermediate fields once all the subgroups are known?
I'm guessing, in general and maybe in this example with $D_8$, there isn't a good, canonical way to anticipate and construct the field extensions? The structure of groups and subgroups, as stated earlier, is easier and better understood than the structure of field extensions. Maybe this makes sense because the groups are finite and have only one operation, and fields are often infinite and have two operations.

and

UPDATE: In this lecture, Richard Borcherds clearly describes how to obtain the two nonobvious intermediate fields from the subgroups. Specifically, add the roots $\alpha$ and $\alpha i$ to fix by reflection one way, and then add the roots $\alpha$ and $-\alpha i$ to fix by reflection the other way.