Is Hoffman elimination anti or syn elimination?
There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions, the E1 and E2 pathways can operate in competition with one another. Activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, the relative strength of the nucleophile, the relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.


(Taken from Ron's answer, in order to start an explanation)(1)

From Wikipedia: 


  The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favour an E1cB mechanism.


So on further thinking about the Hofmann elimination, you notice the following 


$\ce{NR_3^+}$ is a poor leaving group (Smith, March. Advanced Organic Chemistry 6th ed. (501-502))
Due to this, the rate of cleavage of the $\ce{C-H}$ bond cleavage exceeds the rate of the $\ce{C-N}$ bond cleavage
In fact, the three well known poor leaving groups namely $\ce{-F}$, $\ce{-SMe^{2+}$, $\ce{NR^{3+}$ all have the same observation 


What's the conclusion?

Of course, these undergo $E1cb$ when heated to a suitable temperature

There are sources which infer the same -  


Wikipedia 
Organic mechanisms by Xiaoping Sun (3)

  Some examples of compounds that contain poor leaving groups and can undergo the E1cB mechanism are alcohols and fluoroalkanes.


  The $\ce{-NMe3}$ group is electronegative, it can activate beta hydrogen to make it acidic, vulnerable to attack by a base. This gives rise to carbanion  character (e1cb) for the transition state.



Hence, $E_1cb$ is the correct answer