I will concentrate on the case of odd $q$ and $\xi,-\eta$ being squares, but the solution should be extendable to the remaining cases. For genus-0 curves we have a polynomial formula for the point count, which can be encapsulated in terms of the following identity: $$(\sim)\,\sum_{x \in \mathbb{F}_q} \chi(x^2+A) = \begin{cases} -1 &\text{ if }A \neq 0,\\ q-1 & \text{ otherwise,}\end{cases}$$ where $\chi$ is the unique unique non-trivial quadratic character of $\mathbb{F}_q^{\times}$, extended to give $0$ on $0$. Luckily, the computation of $N$ reduces to the study of points on genus-0 curves.
Replacing $(b,d)$ by $(b/\sqrt{\xi},d/\sqrt{\xi})$ we see that we may assume that $\xi = 1$. Further replacing $(c,d)$ by $(c\sqrt{-\eta},d\sqrt{-\eta})$ we see that we may assume that $-\eta = 1$.
Let us expand your defining hypersurface and express it as a quadratic polynomial in $a$:
$$(*) \, a^2(d^2-1) + a(-2bcd) + b^2c^2+b^2-c^2+d^2-1 = 0.$$
The case $d^2 =1$ simplifies to $a(2bcd) = b^2c^2+b^2-c^2$. If $bc \neq 0$, this determines $a$ uniquely. If $bc=0$, we must have $b=c=0$, and $a$ can be arbitrary. All in all, this case brings $2((q-1)^2 + q)$ solutions.
Let us assume $d^2 \neq 1$. The discriminant of $(*)$ factorizes as $4(c^2+1-d^2)(b^2+d^2-1)$, which is a very useful coincidence. Hence we see that given $b,c,d$ contribute to $(*)$ $$1+\chi( (c^2+1-d^2)(b^2 +d^2-1) )$$ solutions.
Summarizing, we have $$(**)\, N=2(q^2-q+1) + q^2(q-2) + \sum_{d^2 \neq 1, \, b,c,d \in \mathbb{F}_q} \chi( (c^2+1-d^2)(b^2 +d^2-1) ).$$
By $(\sim)$, $$\sum_{b \in \mathbb{F}_q} \chi( (c^2+1-d^2)(b^2 +d^2-1) ) = \chi(c^2+1-d^2) \sum_{b \in \mathbb{F}_q} \chi(b^2 +d^2-1 ) = -\chi(c^2+1-d^2)$$ when $d^2 \neq 1$, and so $$\sum_{d^2 \neq 1, \, b,c,d \in \mathbb{F}_q} \chi( (c^2+1-d^2)(b^2 +d^2-1) )= -\sum_{d^2 \neq 1, \, c,d \in \mathbb{F}_q} \chi(c^2+1-d^2),$$ and applying $(\sim)$ once more this becomes $$-\sum_{d^2 \neq 1, \, c,d \in \mathbb{F}_q} \chi(c^2+1-d^2) = \sum_{d^2 \neq 1} 1 = (q-2).$$
Plugging this character sum evaluation in $(**)$, $q^3-q$ solutions are obtained, confirming your empirical observation.
For general $\xi,-\eta$, a very similar argument will work, because the discriminant of your defining equations (considered as a quadratic polynomial in $a$) still factorizes nicely, specifically it is $$4(c^2 - \eta-d^2\xi) (-b^2\xi\eta + d^2 \xi + \eta).$$
Even $q$ is easier. The defining equation can now be written as $(a+1)^2 (\xi d^2 +\eta) = \xi b^2 (c^2+\eta) + c^2 = 0$. In $\mathbb{F}_q$ with even $q$, $x \mapsto x^2$ is a field automorphism.
- If $\xi d^2+\eta \neq 0$ (happens for all but a unique $d$), $a+1$ is uniquely determined by $b,c$, giving $(q-1)q^2$ solutions.
- If $\xi d^2+\eta = 0$, $d$ is determined uniquely, $a$ is arbitrary and it remains to count solutions $(b,c)$ to $c^2(\xi b^2+1) = \eta \xi b^2$. Specifying $b$ determines $c$ uniquely, unless $\xi b^2 + 1=0$ (happens for a unique $b$), in which case there are no solutions. So this case contributes $q(q-1)$ solutions.
All in all, $N = (q-1)q^2+q(q-1) = q^3-q$ for even $q$ as well.