The bond angle of a molecule depends on several factors. We have to look at all the factors and then decide the result according to them.
The difference in bond angles in $\ce{NF3}$ and $\ce{NH3}$ is only determined by the electronegativity difference between the central atom and the bonded atom. As $\ce{F}$ is more electronegative than $\ce{H}$, in $\ce{NH3}$, bonding electron pair shifts more towards $\ce{N}$ than they shift in $\ce{NF3}$. So, the bond angle increases. There is no other effect at all in this case.
But in the case of $\ce{NCl3}$, you can notice there is a lone pair on the central $\ce{N}$ atom. Most importantly, there is an introduction of vacant low lying $\ce{3d}$ orbital of $\ce{Cl}$ where the lone pair of $\ce{N}$ can delocalise. This effect is known as $\ce{p\pi-d\pi}$ backbonding. So, due to this backbonding, which is quite strong, the bond between $\ce{N}$ and $\ce{Cl}$ adopts a significant double bond character.
Although the electronegativity difference is decreased in the case of $\ce{NCl3}$ and bond pair is shifted away from central $\ce{N}$, this introduction of double-bond character increase the repulsion between the bond pairs even more, and that's why the bond angle increases.
So, in case of $\ce{NF3}$ and $\ce{NH3}$ there was no vacant and valence d orbitals either in $\ce{F}$ or in $\ce{H}$. So, this backbonding doesn't occur there. But in $\ce{NCl3}$, the case is different and more factors determine the bond angle. As the backbonding is much stronger, this effect overall dominates over the electronegativity factor and make the bond angle increase.
Experimental evidence of this double bond character: In $\ce{NCl3}$, $\ce{N-Cl}$ bond length is 1.759 angstroms but $\ce{N-Cl}$ single bond length is 1.91 angstroms. This indicates the presence of some partial double bond character which shrinks the bond length.