By symmetry you know that the stress at each plane of symmetry must be perpendicular to the plane. Take two planes of symmetry at 30 degrees from each other to section off 1/12th of the hexagon. Now you just have a beam with a constant pressure and two point loads at the corners.
Since the net force must be zero, the component of the force applied at the hexagon corner that is perpendicular to the beam, must cancel the force of the pressure on that beam.
$$\sin(30 deg)\,F_c=P\,a$$
Where $F_c$ is the magnitude of the force per container length at the corner, $P$ is the pressure, and $a$ is the length of half the side of the hexagon.
Then the forces parallel to the beam must balance:
$$F_m=\cos(30 deg)\,F_c$$
Now we have the beam loading, so we can do a beam analysis:
$$\text{Shear stress}=P\,x$$
$$\text{Bending Moment}=\frac12 P\,x^2+C$$
$$\text{Slope}=\frac1{E\,I}\left(\frac16 P\,x^3 + C\,x + D\right)$$
Now by symmetry we know the slope must be zero at the ends:
$$0=\frac1{E\,I}\left(\frac16 P\,0^3 + C\,0 + D\right)$$ $$0=D$$ $$0=\frac1{E\,I}\left(\frac16 P\,a^3 + C\,a\right)$$ $$C=-\frac16 P\,a^2$$
$$\text{Bending Moment}=P\left(\frac12 x^2-\frac16 a^2\right)$$
Now that we have the tension $F_m$ and the bending moment we can calculate stress on the inner and outer surface of the skin:
$$\sigma=-\frac{F_m}{t}\pm \frac{P}{t^2}\left(\frac12 x^2-\frac16 a^2\right)$$ $$\sigma=-\frac{\cot(30 deg)P\,a}{t}\pm \frac{P}{t^2}\left(\frac12 x^2-\frac16 a^2\right)$$
This has a maximum of: $$\sigma=-P\left(\frac{\sqrt{3}\,a}{t} + \frac{a^2}{3t^2} \right)$$
Now we can compare this to a cylinder as an inscribed circle $a=\frac1{\sqrt{3}}r$
$$\sigma=-P\left(\frac{r}{t} + \frac{r^2}{9t^2} \right)$$
The first term, which comes from the tension in the beam is the same as the stress for the cylinder. 333
The second term, which comes from the bending stress dose not appear in the case of the cylinder as there is no bending. This bending stress will dominate the expression for thin walled vessels to the point were it is likely they would just deform until they resembled a cylinder.
As an added bonus this analysis works for any regular polygon:
$$\sigma=-P\left(\frac{r}{t} + \tan(\alpha)^2\frac{r^2}{3t^2} \right)$$
Where $\alpha$ is half of the central angle of the polygon. As the number of sides increases, $\alpha$ approaches zero as does the bending moment term, converging on the solution for the cylinder. 333