Denote an $n$-dimensional gamma constant as $\gamma_n$.
$\large \textbf{Background}$
This article gives a nice $2D$ graphical representation of Euler's constant.
The area of the blue dots represents the Euler's constant $ 0.577215664...$
Let's revolve the bars to form an infinite stack of cylinders with volume
$$V_c = \frac{\pi^{\frac{n-1}{2}} \zeta(n-1)}{\Gamma\left( \frac{n+1}{2} \right)}$$
Let's revolve the curve $y=\frac{1}{x}$ to form Gabriel's horn with volume $\dagger$
$$V_h = \frac{\pi^{\frac{n-1}{2}}}{(n-2) \Gamma\left( \frac{n+1}{2} \right)}$$
Subtracting $V_c - V_h$ gives the $n$-dimensional Euler-Mascheroni constant for $n \ge 2$
$$\bbox[5px,border:2px solid #0000ff]{\gamma_n = \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left( \frac{n+1}{2} \right)} \left( \zeta(n-1) - \frac{1}{n-2} \right)}$$
where $\Gamma$ is Euler's gamma function.
$\large \textbf{Notes}$
- For $n=2$
$$\gamma_2=2\gamma$$
The Behaviour of $\zeta(s)$ near $s=1$ is$$\zeta\left(s\right)=\frac{1}{s-1}+\gamma+O\left(\left|s-1\right|\right)$$
$$\implies\zeta\left(n-1\right)=\frac{1}{n-2}+\gamma+O\left(\left|n-2\right|\right)$$
$$\implies\zeta\left(n-1\right) - \frac{1}{n-2}=\gamma+O\left(\left|n-2\right|\right)$$
$$\implies\gamma_n = \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left( \frac{n+1}{2} \right)} \left(\gamma+O\left(\left|n-2\right|\right)\right)$$
$$\implies\gamma_2 = \frac{\sqrt{\pi}}{\Gamma\left( \frac{3}{2} \right)}\gamma = \frac{\sqrt{\pi}}{\frac{\sqrt{\pi}}{2}}\gamma = 2\gamma$$
- We can express $\zeta$ in terms of $\gamma_n$
$$\zeta(n-1) = \frac{\Gamma\left( \frac{n+1}{2} \right) \gamma_n}{\pi^{\frac{n-1}{2}}} + \frac{1}{n-2}$$
$\large \textbf{Example 3D gamma constant}$
Set $n = 3$
$$\gamma_3 = \frac{\pi^{\frac{3-1}{2}}}{\Gamma\left( \frac{3+1}{2} \right)} \left( \zeta(3-1) - \frac{1}{3-2} \right)$$
Simplifying
$$\gamma_3 = \frac{\pi}{\Gamma(2)} \left( \zeta(2) - 1 \right)$$
where $\Gamma(2) = 1$ and $\zeta(2) = \frac{\pi^2}{6}$, simplifies to
$$\gamma_3 = \pi \left( \frac{\pi^2}{6} - 1 \right) = 2.026120126...$$
$\large \textbf{Question}$
When extended to higher dimensions, I'm not $100\%$ sure the Gabriel's Horn derivation is accurate?
$\dagger$ I tried the following:
The volume of an $n$-dimensional hypersphere is
$$V_n(R) = \frac{\pi^{n/2} R^n}{\Gamma\left(1 + \frac{n}{2}\right)}$$
and for $n-1$ dimensions with radius $R=\frac{1}{x}$ is
$$V_{n-1}(\frac{1}{x}) = \frac{\pi^{\frac{n-1}{2}}}{x^{n-1}\Gamma\left( \frac{n+1}{2} \right)}$$
So, the total volume of Gabriel's Horn hyperspherical slices along the $x$-axis is
\begin{align*}\int_1^\infty \frac{\pi^{\frac{n-1}{2}}}{x^{n-1} \Gamma\left( \frac{n+1}{2} \right)} \, dx&= \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left( \frac{n+1}{2} \right)} \int_1^\infty \frac{1}{x^{n-1}} \, dx \\&= \frac{\pi^{\frac{n-1}{2}}}{(n-2) \Gamma\left( \frac{n+1}{2} \right)} \quad \blacksquare\end{align*}where$$\int_1^\infty \frac{1}{x^{n-1}} \, dx = \frac{1}{n-2}$$