= Solution
A normalization is important here. I use the <covolume-one theta function of a fractional ideal>, for which the requested limit is $1$, and also give the formula for the unnormalized <Gaussian theta sum>.
Let $n=[K:\mathbb Q]$, let $d_v=[K_v:\mathbb R]\in\{1,2\}$, and choose one <field embedding> $\sigma_v$ for each <Archimedean place>. The <Minkowski embedding of a number field> identifies
$$
K_\infty=\mathbb R^{r_1}\times\mathbb C^{r_2},\qquad n=r_1+2r_2,
$$
with a real <inner-product space> having
$$
\langle x,z\rangle=\sum_{v\text{ real}}x_vz_v+2\sum_{v\text{ complex}}\operatorname{Re}(x_v\overline{z_v}),\qquad |x|^2=\sum_vd_v|x_v|^2.
$$
Its Euclidean <Lebesgue measure> is $d\mu=\prod_{v\text{ real}}dx_v\prod_{v\text{ complex}}2\,d\operatorname{Re}x_v\,d\operatorname{Im}x_v$. With this convention, the <covolume of a fractional ideal lattice> $j(\mathfrak b)$ is
$$
C_{\mathfrak b}=\sqrt{|D_K|}\,N(\mathfrak b),
$$
where $D_K$ is the <field discriminant> and $N(\mathfrak b)$ is the positive absolute <norm of a fractional ideal>. Thus the <Euclidean lattice> $\Lambda_{\mathfrak b}=C_{\mathfrak b}^{-1/n}j(\mathfrak b)$ has <covolume> $1$. Define
$$
\boxed{\Theta(y,\mathfrak b)=\sum_{a\in\mathfrak b}\exp\left(-\pi C_{\mathfrak b}^{-2/n}\sum_{v\in S_\infty}d_vy_v|\sigma_v(a)|^2\right).}
$$
This <Gaussian theta sum> converges absolutely whenever every $y_v>0$.
The <trace dual of a fractional ideal> is
$$
\mathfrak b^\vee=\mathfrak D_{K/\mathbb Q}^{-1}\mathfrak b^{-1}=\{z\in K:\operatorname{Tr}_{K/\mathbb Q}(z\mathfrak b)\subseteq\mathbb Z\}.
$$
Here $\mathfrak D_{K/\mathbb Q}^{-1}$ is the <inverse different>. Since $N(\mathfrak D_{K/\mathbb Q})=|D_K|$, we have $C_{\mathfrak b^\vee}=C_{\mathfrak b}^{-1}$. There is a subtle distinction between the <trace pairing> and the positive <inner product>: the <dual lattice> of $j(\mathfrak b)$ is $\overline{j(\mathfrak b^\vee)}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,
$$
\langle j(a),\overline{j(z)}\rangle=\operatorname{Tr}_{K/\mathbb Q}(az).
$$
Consequently $\Lambda_{\mathfrak b}^*=\overline{\Lambda_{\mathfrak b^\vee}}$. Coordinatewise <complex conjugation> preserves the weighted squared lengths in the <Gaussian theta sum>.
Here are the precise analytic formulas used in the proof. For a <Schwartz function> on $K_\infty$, take the <Fourier transform> to be
$$
\widehat f(z)=\int_{K_\infty}f(x)e^{-2\pi i\langle x,z\rangle}\,d\mu(x).
$$
For a full <Euclidean lattice> $\Lambda$ of <covolume> $C$, the <Poisson summation formula for a Euclidean lattice> is
$$
\sum_{x\in\Lambda}f(x)=C^{-1}\sum_{z\in\Lambda^*}\widehat f(z).
$$
For $f_y(x)=\exp(-\pi\sum_vd_vy_v|x_v|^2)$, the <Gaussian Fourier transform>, applied in orthonormal real coordinates, gives
$$
\widehat f_y(z)=\|y\|^{-1/2}\exp\left(-\pi\sum_vd_vy_v^{-1}|z_v|^2\right),\qquad \|y\|=\prod_vy_v^{d_v}.
$$
More generally, $\widehat{\exp(-\pi x^TAx)}(z)=(\det A)^{-1/2}\exp(-\pi z^TA^{-1}z)$ for a real <positive-definite matrix> $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.
Apply the <Poisson summation formula for a Euclidean lattice> to $\Lambda_{\mathfrak b}$, whose <covolume> is $1$, and use the preceding description of its <dual lattice>. The <anisotropic theta functional equation> is
$$
\boxed{\Theta(y,\mathfrak b)=\|y\|^{-1/2}\Theta(y^{-1},\mathfrak b^\vee),\qquad (y^{-1})_v=y_v^{-1}.}
$$
For comparison, the unnormalized <Gaussian theta sum>
$$
\theta(y,\mathfrak b)=\sum_{a\in\mathfrak b}\exp\left(-\pi\sum_vd_vy_v|\sigma_v(a)|^2\right)
$$
has the <functional equation>
$$
\theta(y,\mathfrak b)=C_{\mathfrak b}^{-1}\|y\|^{-1/2}\theta(y^{-1},\mathfrak b^\vee),
$$
and its corresponding limit is $C_{\mathfrak b}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta(y,\mathfrak b)=\theta(C_{\mathfrak b}^{-2/n}y,\mathfrak b)$.
For the <small-parameter asymptotic of a lattice theta sum>, $y\to0$ means that every coordinate tends to zero. In $\Theta(y^{-1},\mathfrak b^\vee)$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\leq1$, all these terms are bounded by the summable <Gaussian theta sum> $\sum_{\lambda\in\Lambda_{\mathfrak b^\vee}}e^{-\pi|\lambda|^2}$. The <dominated convergence theorem> therefore gives $\Theta(y^{-1},\mathfrak b^\vee)\to1$. Using the <anisotropic theta functional equation>,
$$
\boxed{\lim_{y\to0}\|y\|^{1/2}\Theta(y,\mathfrak b)=1.}
$$
The condition that every coordinate tends to zero matters; $\|y\|\to0$ alone would not justify this argument.
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