Let
$$ \begin{array}{l} f: [0,\frac{\pi}{2}]^3 \to {\mathbb R}^+, \\ (\theta_1,\theta_2,\theta_3) \mapsto |2+e^{i\theta_1}+e^{i\theta_2}+e^{i\theta_3}| \end{array} $$
Numerical values suggest that the minimum of $f$ is $\sqrt{13}$, and is attained at the following four points : the three permutations of $(0,\frac{\pi}{2},\frac{\pi}{2})$, and $(\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2})$. But I was unable so far to show it rigorously, any help appreciated.
$$g(\theta_1,\theta_2,\theta_3):=f^2(\theta_1,\theta_2,\theta_3)$$ then $$g(\theta_1,\theta_2,\theta_3)=|2+e^{i\theta_1}+e^{i\theta_2}+e^{i\theta_3}|^2\\=(2+\cos\theta_1+\cos\theta_2+\cos\theta_3)^2+(\sin\theta_1+\sin\theta_2+\sin\theta_3)^2$$ $g$ is minimum whenever $f$ is minimum. Consider the following optimization problem $$\min_{\theta} g(\theta)\hspace{0.2cm}\text{subject to}\hspace{0.2cm}\theta\in[0,\frac{\pi}{2}]^3$$ where $\theta:=(\theta_1,\theta_2,\theta_3)$. The Lagrangian for this problem is $$L(\theta,\lambda,\mu):=g(\theta)+\sum_{i=1}^3\lambda_i(\frac{\pi}{2}-\theta_i)-\sum_{i=1}^3\mu_i\theta_i$$ where $\lambda,\mu\geqslant 0$. First order conditions wrt to $\theta_1,\theta_2,\theta_3$ are $$-2(2+\cos\theta_1+\cos\theta_2+\cos\theta_3)\sin\theta_i+2(\sin\theta_1+\sin\theta_2+\sin\theta_3)\cos\theta_i-\lambda_i+\mu_i=0$$ for $i=1,2,3$. These equations can be simplified further to $$2\sin\theta_1-\lambda_1+\mu_1=(\sin\theta_2\cos\theta_1-\cos\theta_2\sin\theta_1)+(\sin\theta_3\cos\theta_1-\cos\theta_3\sin\theta_1)\\=\sin(\theta_2-\theta_1)+\sin(\theta_3-\theta_1)$$ Analogue $$2\sin\theta_2-\lambda_2+\mu_2=\sin(\theta_1-\theta_2)+\sin(\theta_3-\theta_2)$$ $$2\sin\theta_3-\lambda_3+\mu_3=\sin(\theta_1-\theta_3)+\sin(\theta_2-\theta_3)$$ Adding all these three equations together and using $\sin(-x)=-\sin x$ yields $$\sin\theta_1+\sin\theta_2+\sin\theta_3=\frac{1}{2}\sum_i(\lambda_i-\mu_i)$$ From slackness conditions $\lambda_i(\pi/2-\theta_i)=0, \mu_i\theta_i=0$ we observe that if the optimal $\theta$ in in the interior of $[0,\pi/2]^3$ then $\lambda_i=\mu_i=0$ for $i=1,2,3$ which would force $$\sin\theta_1+\sin\theta_2+\sin\theta_3=0$$ Since for $\theta\in(0,\pi/2]$ we have $\sin\theta>0$ then it must be the case that the above equation is satisfied only for $\theta_1=\theta_2=\theta_3=0$ contradicting the fact that $\theta$ was an interior point. Therefore the optimal value must lie on the boundary. One can check case by case the boundary values and compare their critical values.