Let $\alpha=j_{n,m}$ be the positive zeros of $J_n(\alpha)$, numbered so that $j_{n,m+1} > j_{n,m}$. Let $m_{n,\alpha}$ denote the largest $m$ such that $j_{n,m} < \alpha$. It seems to me that you want to asymptotically estimate the sum
$$N_\lambda = \sum_{n<\lambda/C} m_{n,Cn}.$$
Since the numbers $m_{n,Cn}$ would be increasing, the asymptotics should be dominated by the largest term $m_{\lambda/C,\lambda}$. To get that, you need to solve the equation $j_{\lambda/C,m} = \lambda$ for $m$ as $\lambda \to \infty$. There is an asymptotic expansion for $j_{n,m}$ for large $n$ that is uniform in $m$ (DLMF, §10.21.41), which is what I think should be used here. If I have decoded this uniform approximation correctly, then
$$ m_{\lambda/C,\lambda} \sim K \lambda, \quad \text{with} \quad
K = \frac{\sqrt{C^2-1}-\sec^{-1}C}{\pi C} . $$
So your estimate should be $N_\lambda \sim K \lambda^2/2$.