$\newcommand\Q{\mathbf{Q}}
\newcommand\OL{\mathcal{O}}
\newcommand\I{\mathcal{I}}
\newcommand\Z{\mathbf{Z}}
\newcommand\eps{\epsilon}
\newcommand\p{\mathfrak{p}}
\newcommand\PP{\mathfrak{P}}
\newcommand\Hom{\mathrm{Hom}}
\newcommand\R{\mathbf{R}}
\newcommand\q{\mathfrak{q}}
\newcommand\Gal{\mathrm{Gal}}$
Summary: I think that it's difficult problem and that there won't be a "nice" answer in general.
Let $\I_K$ denote the group of invertible fractional ideals.
There is a tautological exact sequence
$$1 \rightarrow K^{\times}/\OL^{\times} \rightarrow \I_K
\rightarrow C_K \rightarrow 0.$$
Taking cohomology gives the following sequence:
$$0 \rightarrow (K^{\times}/\OL^{\times})^{G}
\rightarrow \I^{G}_K
\rightarrow C^{G}_K \rightarrow H^1(G,K^{\times}/\OL^{\times}),$$
which can naturally be modified to yeild the sequence:
$$0 \rightarrow (K^{\times}/\OL^{\times})^{G}/\Q^{ > 0}
\rightarrow \I^{G}_K/\Q^{ > 0}
\rightarrow C^{G}_K \rightarrow H^1(G,K^{\times}/\OL^{\times}).$$
You are interested in the order of $ (K^{\times}/\OL^{\times})^{G}/\Q^{ > 0}$.
As Will Sawin noted, the second term is simply $\prod \Z/e_p$. Thus one wants
to understand the classes $I \in C^G_K$ (that is, the so-called ambiguous
classes)
which are actually strongly ambiguous , that is, $\sigma I = I$ as ideals , not
just ideal classes. If one defines $S_K \subset C^G_K \subset C_K$ to denote the subset of strongly
ambiguous classes, then one "formally" has an answer to your question, namely,
$$\frac{1}{|S_K|} \cdot \prod e_p.$$
OTOH, this is really a proof by definition, so much content so far, although it gives a "name" to some of the objects to connect you with the literature.
The strongly ambiguous classes can also be described (given the exact sequence above)
as the kernel of the map
$$C^G_K \rightarrow H^1(G,K^{\times}/\OL^{\times})
\hookrightarrow H^2(G,\OL^{\times})$$
What makes things much easier when $K$ is cyclic is that one can essentially determine
$C^G_K$ (by the ambiguous class number formula, which only exists for cyclic extensions), and also $S_K$. In fact, for a cyclic extension
$S_K$ and $C^G_K$ are (almost) the same group.
Let $C^{+}_K$ denote the group of invertible fractional ideals $I_K$ modulo the following relation:
$[I] \sim [J]$ if and only if $I = (\alpha) J$ for some $\alpha$ with $N(\alpha) > 0$. In particular, $C^{+}_K$
is a quotient of the narrow class group, and surjects onto the class group. For example, $C^{+}_K = C_K$
if there exist units of norm $-1$. Suppose that $K/\Q$ is cyclic. I claim that the image of $(C^{+}_K)^G$
in $C^{G}_K$ lands in $S_K$. By assumption, $[\sigma I] \sim [I]$ in $C^{+}_K$, so $\sigma I = (\alpha) I$ for some $\alpha$ with positive norm.
Clearly $N(\alpha)$ is a unit, so $N(\alpha) = 1$. Since $K/\Q$ is cyclic, by Hilbert 90 there exists a $\beta \in K^{\times}$
such that $\alpha = \beta/\sigma \beta$. (This version of Hilbert 90 only makes sense for cyclic extensions, which is one
of the difficulties in the general case.) Replacing $I$ by $J = (\beta) I$ we deduce that $\sigma J = J$ and $[I] = [J]$ in
$C_K$. When $K/\Q$ is not cyclic, however, then it's much trickier to get a handle on the
ambiguous classes --- I don't think that there will be a nice answer in general.
The simplest possible non-cyclic case is the case of biquadratic extensions. Suppose that $K$ is totally real.
Then $K$ contains three subfields $K_1$, $K_2$, and $K_3$. The unit group of $K$ is, up to finite index,
generated by $U:=\{\eps_1, \eps_2, \eps_3\}$. Kubota
(Uber den bizyklischen biquadratischen Zahlkörper) classified the possible $\Gal(K/\Q)$-module
structure of $\OL^{\times}_K$ (there are $8$ or so different types, of indices ranging from $2^0$ to $2^3$). One could simply compute the corresponding
cohomology groups of each type, and see what one gets.
It's not clear that the answer will be any more precise than a list of cases.
(Even in the case of real quadratic fields the answer depends on the existence of unit of norm
$-1$, which is itself a notoriously fickle condition.)
Yet another way to explain why the cyclic case is not typical is that, since (at least for $p$ odd) the units tensor $\Z_p$ are annihilated by the norm map, they form a module under $$\Z_p[x]/(x^n-1,1+x+x^2+ \ldots + x^{n-1}) = \Z_p[\zeta_n],$$
which essentially a direct sum of PIDs.
I guess it depends on exactly what you are interested in doing, but it might be useful
to consider the following approach. Assuming $K/\Q$ is abelian, one can easily compute
the genus class field $L/K$. The field $L$ is the largest field such that $L/\Q$ is totally real, abelian, and $L/K$ is unramified everwhere.
The point of working with $L$ is that one knows that
the strongly ambiguous classes in $\OL_K$ (and in $\OL_L$) become (are) principal in $\OL_L$, so the transgression map
is injective, and
$$(L^{\times}/\OL^{\times}_{L})^{G_L}/\Q^{ > 0} \simeq
\I^{G}_L/\Q^{ > 0} \simeq \prod \Z/e_p.$$
(The $e_p$ are the same for $L$ and $K$ since $L/K$ is everywhere unramified.)
If $L = \Q(\zeta_N)$, then one has explicit generators for this group, namely,
$1 - \zeta$ for each $p^n$th root of unity for $p^n \| N$.
Note that, in some sense, this gives a "complete" description of the $d$-integers in $\Q^{\mathrm{ab}}$:
The $d$-numbers in $K$ are
the $G_K$-invariants of the finitely generated group
$$\Z[\zeta_N]^{\times} \times \prod_{p^n \| N} (1 - \zeta)^{\Z}.$$
Concerning the image of the norm map, I think you are again out of luck.
To explain why, consider what is close to the simplest possible non-trivial example, namely,
$K = \Q(\sqrt{6p})$, where
$p \equiv 1 \mod 4$ is prime. Write $(2) = \p^2_2$, $(3) = \p^2_3$, and
$(p) = \p^2_p$ respectively.
The field $K$ does not contain a unit of norm $-1$, so we know that the
$d$-numbers generate the group $(\Z/2\Z)^2$. Since $\p_2 \p_3 \p_p$ is principal, it follows
that exactly one of $\p_2$,
$\p_3$, and $\p_p$ will be principal.
The genus class field of $K$ is
$$F = \Q(\sqrt{6},\sqrt{p}).$$
It follows that if $C_K$ is the class group of $K$, then
$C_K/2C_K$ is cyclic, and hence the $2$-part of the class group is
cyclic. Via the Artin map, we can identify the primes ideals which lie in $C_K[2]$ as exactly the primes which
split completely in the genus field. Since the genus field is given explicitly, we may compute that
- $\p_2$ splits in $F/K$ if and only if $p \equiv 1,17 \mod 24$,
- $\p_3$ splits in $F/K$ if and only if $p \equiv 1,13 \mod 24$,
- $\p_p$ splits in $F/K$ if and only if $p \equiv 1,5 \mod 24$.
In particular, if $p$ lies outside one of these equivalence classes, then
the image of the corresponding $\p$ is non-trivial in $C_K/2 C_K$, and hence $\p$ is not principal.
It follows that:
- If $p \equiv 17 \mod 24$, then the $d$-numbers are generated by $2$ and $6p$,
- If $p \equiv 13 \mod 24$, then the $d$-numbers are generated by $3$ and $6p$,
- If $p \equiv 5 \mod 24$, then the $d$-numbers are generated by $6$ and $6p$,
This leaves open the case when
$q \equiv 1 \mod 24$. The computation above merely shows that
$\p_2$, $\p_3$, and $\p_p$ all become trivial in $C_K/2C_K$. Since they cannot all be principal,
it follows that when $p \equiv 1 \mod 4$, there must be a surjection
$C_K \rightarrow \Z/4\Z$. By class field theory, this corresponds to the existence of an unramified extension
$E/K$ with $\Gal(E/K) = \Z/4\Z$. We may construct $E$ explicitly as follows.
Since $\Z[\sqrt{6}]$ has class number one, and $p \equiv 1 \mod 24$ splits this field, there exists
a $\pi \in \Z[\sqrt{6}]$ with $N(\pi) = p$. (For local reasons the sign will be positive.)
The choice of $\pi$ will be unique up to a sign and the fundamental unit $\eps = 5 + 2 \sqrt{6}$. If we explicitly write
$\pi = A + B \sqrt{6}$, then we have
$$A^2 - 6 B^2 = p.$$
The congruence on $p$ forces $B$ to be even and $A$ to be odd. After possibly multiplying by a unit and by $-1$, we
may assume that $A \equiv 1 \mod 4$ and $B \equiv 0 \mod 4$. This determines $\pi$ up to squares, and $E$
is identified with the Galois closure of $\Q(\sqrt{\pi}) = \Q(\sqrt{A + B \sqrt{6}})$.
In particular, $E$ is the splitting field of
$$X^4 - 2 A X^2 + p.$$
A necessary condition (and sufficient if the $2$-part of the class group has order $4$) for $\p_p$ to be prime is
that the residue degree of $p$ in $E$ is $1$, or equivalently that $(2 A/p) = 1$.
Since $p \equiv 1 \mod 24$, this is the same as saying that $(A/p) = 1$. In fact, (proof omitted, because this
answer is already too long and the argument is a somewhat tedious calculation of ring class fields and Kummer extensions)
this is equvalent to $(6/p)_4 = 1$. So this leads to the following criterion:
If $p \equiv 1 \mod 24$, and the quartic residue $(6/p)_4 = -1$, then there are no $d$-numbers of norm $p$.
If
$p \equiv 1 \mod 24$, and $8 \nmid h_K$, then there do exist $d$-numbers of norm $p$.
One could go on, giving similar criteria for $\p_2$ and $\p_3$, but it will just get worse (at least for $4 \| h_K$
one can give some sort of classical criteria due to the existence of governing fields, probably for
$8 \| h_K$ and certainly for $16 \| h_K$ there won't be any non-tautological criterion.)
All in all, the "best" cases are when $K$ is its own genus field, or at least, for all $p$ dividing $[K:\Q]$, the $p$-class field of $K$ is the $p$-genus field of $K$.
If this doesn't help, perhaps you could say more precisely you want to prove about dimensions (or otherwise) of fusion categories?
