Angular Momentum in a Straight Line Edit: This is not a duplicate question.
The other question asked how angular momentum remained constant if the distance varied.
This question asks why you can select any point and calculate angular momentum from there, instead of intuitively choosing to calculate the angular momentum with reference to the centre of mass/pivot.  
A box is moving with constant velocity in a straight line. (The box is not rotating about its centre of mass)  
But apparently, you can set the axis of rotation at any point, and the box will have an angular momentum of r x p (r is perpendicular distance from axis of rotation, p is momentum)  
But why can you select the axis of rotation at any point instead of only at a pivot/centre of mass?
 A: Short answer: you can calculate the angular momentum from anywhere you want, as long as the vectors $r$ and $p$ are defined. If what you calculate is useful of easy is another issue, but nothing prevents you from calculating a vector product of two things.
A: You can choose the point about which you calculate angular momentum, just as you can choose the point about which you take moments in a statics set-up (such as an unevenly loaded plank resting on trestles). 
Indeed for a particle displaced by $\vec r$ from a point O, the rate of change of angular momentum, $\vec J,$ about that point is$$\frac{d \vec J}{dt}=\frac{d[\vec r \times (m\vec v)]}{dt}=m\left(\frac{d\vec r}{dt}\times \vec v\ + \vec r \times \frac{d \vec v}{dt}\right)= \vec r \times \frac{d(m\vec v)}{dt}=\vec r \times \vec F$$
[The first term in the big brackets is the cross product of a vector by itself and is therefore zero.] So we have established the important result that
Rate of change of particle's angular momentum about O = Moment of force on particle about point O.  
This applies for any point O that we choose, making it a powerful principle because of its generality!
Although we can choose any point O, there may be strategic reasons for choosing a particular point. For example, if we choose the centre of the Sun as O, then a planet has a fixed angular momentum about that point, because there is no moment of the Sun's force on the planet about that point, as the force is radially inwards towards that point (to a good approximation). The constancy of angular momentum about O 'explains' Kepler's observed equal area law.
