Having scalar product bounded below is equivalent to being at spherical distance bounded below, independently of $d$ (by $\theta=\arccos(\rho)$). Suppose that you have points $x_1,\ldots,x_N$ on the surface of the $(d-1)$-sphere, at spherical distance at least $\theta$ from each other. Then the following open cones $C_i$ are disjoint and all lie in the unit sphere. The intersection of the cone shell $\lbrace y:\langle x_i,y\rangle=\cos(\theta/2)\rbrace$ with the unit sphere is a $(d-2)$-sphere that bounds a $(d-1)$-ball of radius $\sin(\theta/2)$. Then $C_i$ is the cone based on this ball with vertex $0$. The axis of the cone $C_i$ is a segment from the origin in the direction of $x_i$, of length $\cos(\theta/2)$. The cones are disjoint because the angle between a point in $C_i$ and $x_i$ is less than $\theta/2$, while the one between $x_i$ and $x_j$ is at least $\theta$ for $i\neq j$.

This means that $N\cdot|C_1|_d\leq |\mathbb B^d|_d$, where $|\cdot|_d$ denotes the Lebesgue volume in dimension $d$ and $\mathbb B^d$ is the unit ball in $\mathbb R^d$. But the volume of the cone is
$$\frac1d|\mathbb B^{d-1}|_{d-1}\cdot\cos(\theta/2)^{d-1}\cdot\sin(\theta/2).$$
It means that $N$ is at most
$$\frac{|\mathbb B^d|_d}{|\mathbb B^{d-1}|_{d-1}}\cdot\frac{d}{\sin(\theta/2)}\cdot\cos(\theta/2)^{1-d}.$$
Since the first ratio goes to zero (I think as $1/\sqrt d$, up to constant), $N$ has to be asymptotically less than $\lambda^d$ for all $\lambda>1/\cos(\theta/2)$.