Let me improve my comment above. Let $G$ be a finite $p$-group generated by $d$ elements. Let $\Phi(G)=G^p[G,G]$ be its Frattini subgroup. Then $G/\Phi(G)$ is an elementary abelian $p$-group of order $p^d$, that is, a vector space of dimension $d$ over a field of $p$ elements. The number of subspaces in such vector space is about $p^{d^2/4}$ (see somewhere in Chapter 1 of Subgroup Growth by Lubotzky and Segal). Each such subspace corresponds to distinct normal subgroup of $G$. Thus, if $n(G)$ is the number of normal subgroups of $G$, we get that $n(G) \geq p^{d^2/4}$ or in other words $2\sqrt{\log_p(n(G))} \geq d$.

Notice that the best bound is achieved for elementary abelian $p$-group. It seems natural to ask what can be done for finite soluble groups and more generally for any finite group.

Let me improve my comment above. Let $G$ be a finite $p$-group generated by $d$ elements. Let $\Phi(G)=G^p[G,G]$ be its Frattini subgroup. Then $G/\Phi(G)$ is an elementary abelian $p$-group of order $p^d$, that is, a vector space of dimension $d$ over a field of $p$ elements. The number of subspaces in such vector space is about $p^{d^2/4}$ (see somewhere in Chapter 1 of Subgroup Growth by Lubotzky and Segal). Each such subspace corresponds to distinct normal subgroup of $G$. Thus, if $n(G)$ is the number of normal subgroups of $G$, we get that $n(G) \geq p^{d^2/4}$ or in other words $2\sqrt{\log_p(n(G))} \geq d$.

Notice that the best bound is achieved for elementary abelian $p$-group. It seems natural to ask what can be done for finite soluble groups and more generally for any finite group.