Fair warning that
- my algebraic number theory is a bit rusty, and
- this solely reduces your problem to a (standard) problem, namely showing a lower bound on the shortest vector of a random ideal lattice.
Recall that an ideal lattice in some number field $K$ is a lattice that takes the form $L = I\mathcal{O}_K$, i.e. is an ideal $I$ in the ring of integers of $K$.
Ideal lattices, as lattices, satisfy Minkowski's first bound, namely that (in dimension $n$) $\lambda_1(L) \leq \sqrt{n} \sqrt[n]{\det L}$.
Now, given an ideal lattice $I\mathcal{O}_K$, noting that $|N(I)| = |\mathcal{O}_K / I| = \left|\frac{\det I}{\det \mathcal{O}_K}\right|$, we get that
$$\lambda_1(I\mathcal{O}_K) \leq \sqrt{n}\sqrt[n]{\det I}\implies \frac{\lambda_1(I\mathcal{O}_K)}{\sqrt{n}\sqrt[n]{\det \mathcal{O}_K}} \leq \sqrt[n]{N(I)}.$$
This reduces your problem to showing that, for average ideals $I$ (or perhaps principle ideals $(a)$), $\lambda_1(I\mathcal{O}_K)$ is large.
I know of good references for showing that there exists $I$ such that $\lambda_1(I\mathcal{O}_K)$ is large (namely section 4 of this).
A typical reference for discussion that on average integer lattices have $\lambda_1(\mathcal{L}(A))$ large is this or this.
I will also briefly give a (directly) proof that $\lambda_1(\mathcal{L}(A))$ is large on average (for uniform $A$).
The ones in the references use fairly high-powered tools, namely things like Siegal's integration formula.
The direct proof hopefully will show where things break down in the ideal case.
Let $\mathcal{B}_n(c)$ be a (zero-centered) ball of integer vectors of radius at most $c$.
Let $\mathcal{L}(A)$ be the lattice generated by a matrix $A$, which we will assume is uniformly random.
Then, we have that
\begin{align*}
\Pr_A[\lambda_1(\mathcal{L}(A)) > c] &= 1 - \Pr_A[\lambda_1(\mathcal{L}(A)) \leq c]\\
& 1 - \Pr_A[\mathcal{L}(A) \cap \mathcal{B}(c) = \{0\}]\\
&\geq 1 - |\mathcal{B}(c)\setminus\{0\}| \max_{z\in\mathcal{B}(c)\setminus\{0\}}\Pr_A[z \in\mathcal{L}(A)].
\end{align*}
All that we have done now is rearrange things, and apply the union bound.
All that remains is to
- estimate $|\mathcal{B}(c)\setminus\{0\}|$, and
- compute $\Pr_A[z \in\mathcal{L}(A)]$.
The first is a standard (mathematical) computation, see claim 8.2.
The second is also standard (at least for integer lattices) --- one has that
$$\max_{z\neq 0}\Pr_A[z \in\mathcal{L}(A)] = \frac{|\mathcal{L}(A)\setminus \{0\}|}{|\mathbb{Z}^n/q\mathbb{Z}^n\setminus\{0\}|} = \frac{q^n-1}{\det \mathcal{L}(A) - 1}.$$
For integer lattices, it is straightforward to compute $\det \mathcal{L}(A)$ (see for example this).
But this is (roughly) what we are trying to lower bound in the ideal case, i.e. we reduced lower bounding $N(a)$ to lower bounding $\lambda_1(a\mathcal{O}_K)$ to lower bounding $\det a\mathcal{O}_K \approx N(a)\dots$
