Note: This answer comes after a comment in a related post, 2819799, although it was already "green" as i started. Well, the dialog is important, this is my only interest...
Let us denote by
$$t=\sqrt[4]2$$
the generator of the subfield
$$K=\Bbb Q(t)=\Bbb Q(\sqrt[4]2)
$$
of the given splitting field,
$$
E=K(i)=\Bbb Q(\sqrt[4]2,i)\ ,
$$
and since we will write this $t$ very often, so a short-hand was imperative above.
We will also identify it (agains my rules of always working algebraically) with the only complex root of
the polynomial $X^4-2$, which is real, and positive.
The conjugates of $t$ in $\Bbb C$ are
$$
t_1=t\ ,\
t_2=it\ ,\
t_3=i^2t\ ,\
t_4=i^3t\ ,
$$
and corresponding to them we get embeedings $\sigma_j:\Bbb Q(t)\to\Bbb C$, $t\to t_j$.
The system
$$
\{\ 1,\ t,\ t^2,\ t^3\ \}
$$
is a base of $K=\Bbb Q(t)$ over $\Bbb Q$, and computing the trace and the norm
for the extension $K:\Bbb Q$ can be done in terms of it.
The trace is linear, so for computing it, it is enough to compute
$$
\begin{aligned}
\operatorname{Trace}(1) &= \sum_{1\le 4\le j}\sigma_j(1) &&=1+1+1+1 &&=4\ ,\\
\operatorname{Trace}(t) &= \sum_{1\le 4\le j}\sigma_j(t) &&=t+it+-t-it&&=0\ ,\\
\operatorname{Trace}(t^2) &= \sum_{1\le 4\le j}\sigma_j(t^2) &&=t^2-t^2+t^2-t^2 &&=0\ ,\\
\operatorname{Trace}(t^3) &= \sum_{1\le 4\le j}\sigma_j(t^3) &&=t^3-it^3-t^3+it^3 &&=0\ .
\end{aligned}
$$
So for a general
$$
\xi = a+bt+ct^2+dt^3\in K=\Bbb Q(t)\ ,\qquad a,b,c,d\in \Bbb Q\ ,
$$
we have
$$
\operatorname{Trace}_{K:\Bbb Q}\xi = 4a\ .
$$
The norm is relatively simple, too.
First, we need an identity:
$$
\begin{aligned}
&(X+Y+Z+T)
(X+iY-Z-iT)
(X-Y+Z-T)
(X-iY-Z+iT)
\\[3mm]
&\qquad=
\Big(\ (X+Z)^2-(Y+T)^2\ \Big)
\Big(\ (X-Z)^2+(Y-T)^2\ \Big)
\\[3mm]
&\qquad=
\Big(\ (X^2+Z^2-2YT)-(Y^2+T^2-2XZ)\ \Big)
\\
&\qquad\ \phantom{=}
\Big(\ (X^2+Z^2-2YT)+(Y^2+T^2-2XZ)\ \Big)
\\[3mm]
&\qquad=
(X^2+Z^2-2YT)^2-(Y^2+T^2-2XZ)^2
\\[3mm]
&\qquad=
X^4-Y^4+Z^4-T^4
\\
&\qquad\qquad -2X^2Z^2 + 2Y^2T^2
\\
&\qquad\qquad\qquad - 4 X^2 YT + 4Y^2XZ - 4 Z^2 YT + 4T^2XZ\ .
\end{aligned}
$$
Now we can simply compute the norm for the "smaller extension" of an element of the
form $\xi=a+bt+ct^2+dt^3$
$$
\begin{aligned}
\operatorname{Norm}_{K:\Bbb Q}\xi
&=
\operatorname{Norm}_{K:\Bbb Q}(a+bt+ct^2+dt^3)
\\[3mm]
&=
(a+bt+ct^2+dt^3)
(a+ibt-ct^2-idt^3)
(a-bt+ct^2-dt^3)
(a-ibt-ct^2+idt^3)
\\[3mm]
&=a^4-b^4t^4+c^4t^8-d^4t^{12}\\
&\qquad -2a^2c^2t^4 + 2b^2d^2t^8\\
&\qquad\qquad-4a^2bdt^4 + 4b^2act^4-4c^2bdt^8+4d^2act^8
\\[3mm]
&=a^4-2b^4+4c^4-8d^4\\
&\qquad -4a^2c^2 + 8b^2d^2\\
&\qquad\qquad-8a^2bd + 8b^2ac - 16c^2bd + 16d^2ac\ .
\end{aligned}
$$
We can ask the computer for a confirmation of the above:
var('a,b,c,d')
t = 2^(1/4)
prod( [ a + b*i^k*t + c*i^(2*k)*t^2 + d*i^(3*k)*t^3 for k in [0,1,2,3] ] ).expand()
latex(_)
and we get the result:
a^4 - 2*b^4 + 8*a*b^2*c - 4*a^2*c^2 + 4*c^4 - 8*a^2*b*d
- 16*b*c^2*d + 8*b^2*d^2 + 16*a*c*d^2 - 8*d^4
(Result was manually broken in two lines.)
In LaTeX:
$$
a^{4} - 2 \, b^{4} + 8 \, a b^{2} c - 4 \, a^{2} c^{2} + 4 \, c^{4} - 8 \, a^{2} b d - 16 \, b c^{2} d + 8 \, b^{2} d^{2} + 16 \, a c d^{2} - 8 \, d^{4}
$$
Now each element in $E=K(i)$ can be written uniquely in the form
$$
\xi+i\xi'\ .
$$
We can use the above computation and the "transitivity of the related norms".
The norm for the extension $E:K$ for the above element is
$$
(\xi+i\xi')(\xi-i\xi')\ .
$$
This brings in scene $\xi^2$ and ${\xi'}^2$.
We have to isolate the many components of them w.r.t. the basis $1,t,t^2,t^3$,
then apply the above formula.
(For the trace, the game is simpler, the trace over $E:K$ of $\xi+i\xi'$ is the sum of the elements
$\xi\pm i\xi'$, so it is $2\xi$. From here we apply the already computed trace.
It is hard to say more in detail for "generic elements".
