Notation: write $ A \approx B$ for $(A\ll B) \wedge (B\ll A)$.

It's very tempting to apply Poisson sum, so we set up the following formalism:

We'll apply the identity (Poisson summation formula)
$$\sum_{\ell\in\mathbb{Z}} f(\ell) = \sum_{k\in\mathbb{Z}} \hat{f}(k)$$ to the Gaussian $f(x) = \exp(-zx^2)$ with Fourier transform $\hat{f}(k) = \sqrt{\frac{\pi}{z}}\exp (-\frac{4\pi^2}{z}k^2)$. This will allow us to control the blowup in the original equation of the sum over $\underline{\ell}$ when $v$ is close to zero.

Accordingly let $\Theta(z)=\sum_{\ell\in\mathbb{Z}} f(\ell) = \sum_{\ell\in\mathbb{Z}} \exp(-z\ell ^2)$. Our identity then reads
$$\Theta(z) = \sqrt\frac{\pi}{z}\Theta\left(\frac{4\pi^2}{z}\right)$$

Now for an integer $d$ let
\begin{align*} S_d(k) & = \sum_{\underline{\ell}\in\mathbb{Z}^d}\sum_{\gamma>0} \gamma^{-k-3/2}\int_0^\gamma e^{-N\Vert\underline{\ell}\Vert^2v^2/\gamma^2}e^{-v}v^{k+\beta}dv \\
& = \sum_{\gamma>0} \gamma^{-k-3/2}\int_0^\gamma \left(\Theta\left(Nv^2/\gamma^2\right)\right)^d e^{-v} v^{k+\beta}dv\,.
\end{align*}

We will show that $S_d(k)$ converges iff $k>d-1/2$.

To begin, apply the identity above and get
\begin{align*} S_d(k) & = \sum_{\gamma>0} \gamma^{-k-3/2} \int_0^\gamma \left(\frac{\pi\gamma^2}{Nv^2}\right)^{d/2} \left(\Theta\left(\frac{4\pi^2\gamma^2}{Nv^2}\right)\right)^d e^{-v} v^{k+\beta}dv\\
& \approx \sum_{\gamma>0} \gamma^{d-k-3/2} \int_0^\gamma \left(\Theta\left(\frac{4\pi^2\gamma^2}{Nv^2}\right)\right)^d e^{-v} v^{k+\beta}dv\\
\end{align*}

Having extracted the main terms $\gamma^d v^{-d}$, we have $\left(\Theta\left(\frac{4\pi^2\gamma^2}{Nv^2}\right)\right)^d \approx 1$, since $\Theta(z)$ is continuous on $\left[\frac{4\pi^2}{N},\infty\right)$, is at least $1$ everywhere, and satisfies $\lim_{z\to\infty}\Theta(z) = 1$.

It follows that
$$S_d \approx \sum_{\gamma>0} \gamma^{d-k-3/2} \int_0^\gamma e^{-v} v^{k+\beta-d}dv \,.$$

We consider divergence first, using only the zeroes with $\beta = 1/2$. We must then have $k-d+1/2 > -1$ or the integral would diverge at $v=0$. Assume then that $k > d-3/2$. Then $\lim_{\gamma\to\infty} \int_0^\gamma e^{-v} v^{k+\beta-d}dv = \Gamma(k-d+3/2)$ and hence the integral is $\gg 1$. We conclude that
$$S_d(k) \gg \sum_{\beta=1/2} \gamma^{d-k-3/2} \gg \int_1^\infty \gamma^{d-k-3/2}\log \gamma d\gamma$$
which diverges when $d-k-3/2\geq -1$, that is when $k\leq d-1/2$, since a positive proportion of the zeroes are on the line.

For convergence when $k> d-1/2$ we extend the integral to infinity getting
$$S_k(d) \ll \sum_{\gamma>0} \gamma^{d-k-3/2} \Gamma(k-d+\beta+1)\,.$$
Since $k-d+1>1/2$ and $0<beta<1$, the numbers $\Gamma(k-d+\beta+1)$ are uniformly bounded and since $d-k-3/2 < -1$ the sum over the zeroes converges.

Finally, the four times the indicated sum is $S_2(k) - 2 S_1(k) + 4 S_0(k)$. For $k\in[1,3/2]$, the sums $S_1(k),S_0(k)$ converge while $S_2(k)$ diverges so the indicated sum must diverge.