You are thinking that a section at the tip of a wing will behave like a 2D airfoil. This somewhat happens at the center line due to symmetry (and distance from the tips), but the opposite is true at the wing tips.
Lift is the integral of the local load across the span of a wing.
$L = 0.5\,\rho\,V^2\,\int_{-b/2}^{b/2} c\,c_l\,dy$
But that includes the local lift coefficient, which is an integral of pressure across the chord...
$c_l = \frac{1}{c} \int_0^c \left(C_{p,l}-C_{p,u}\right) dx$
Where $C_{p,l}$ and $C_{p,u}$ are the pressure coefficient on the upper and lower surface of a wing.
We can combined these two integrals together to think of Lift as an integral of the difference in pressure coefficient from the top to bottom of the wing over the entire wing area.
$L = 0.5\,\rho\,V^2\,\int_{-b/2}^{b/2} \int_0^c \left(C_{p,l}-C_{p,u}\right) dx\,dy$
This makes the most sense if you think of a wing as a thin camber surface (like from thin airfoil theory). This is very much the case when you use VLM or LLT theory (two options in XFLR5), but is a little bit of an abstraction if you're using a thick-surface theory like a panel code or CFD.
In those cases, things are a little more complex, but it still works out pretty much the same. For now, think of a wing like a 3D version of thin airfoil theory -- a thin camber surface plus a symmetrical thickness form plus a flat plate at an angle of attack. Remember the thickness form does not contribute to lift or moment, so we're safe to ignore it here.
Shrink yourself down to that differential element $dx\,dy$.
If you were to walk a line at the wing tip, you would find that you are in a place where pressure and a surface doesn't work like normal. One infinitesimal increment outboard (off of the wingtip) and you clearly have no surface to push on, so the air pressure does not contribute to the force/moment of the wing. However, without a surface to push on, the lower and upper pressure coefficients are equal $C_{p,l}=C_{p,u}$.
One infinitesimal increment inboard (into the interior of the wing) and you clearly contribute to the integral like normal. The surface of the wing can support the pressure difference that you integrate up to become a lift force.
However, when you are exactly at the wing tip, what happens to $(C_{p,l}-C_{p,u})$? Is there a surface to support a pressure difference? If there was a pressure difference on that edge, what would happen (wouldn't the air just escape around the edge? To the point that there would be no pressure difference.
As it works out, aerodynamics requires that the upper and lower surface pressures are equal exactly at the wingtip. Consequently, that exact line can not contribute to the forces and moments.
Although I've constructed this argument thinking about lift (not moment), the equation is the same -- except the integral includes a moment arm as a distance from the reference point (typically c/4) to the point of integration (x,y). This term would look something like $(x-c/4)$ or maybe $(c/4-x)$ multiplied by the $(C_{p,l}-C_{p,u})$ term. This contribution to pitching moment will still go to zero because $(C_{p,l}-C_{p,u})$ always goes to zero at the tip.
Check out the distribution of spanwise induced drag coefficient. It too will go to zero at the tip.
So, like I started out -- flow at the centerline of a wing behaves very similar to a 2D wing. However, at the tip, flow is so dominated by 3D effects (the presence of the tip) that it can't even support a pressure difference from top to bottom.
