Can negative (abnormally) lattice enthalpies during the formation of ionic compounds exist?
Short answer: Not for "normal" compounds.

First off, let's consider what the lattice enthalpy means. As you've written it, these lattice enthalpy is part of the Born-Haber cycle of formation of an ionic solid from a metal and halogen.

As you see from your table, the energies decrease with:


Small charges (i.e., these are higher for $\ce{Al2O3}$ or $\ce{CaO}$ because of higher ionic formal charges)
Large inter-atomic distances (i.e., increased ion distance between $\ce{Cs}$ and $\ce{I}$ yield lower electrostatic attraction).


The Born-Lande and Kaputinskii equations give some numeric estimates, e.g. (Born-Landé)


  $E = \frac{N_AMz^+z^- e^2 }{4 \pi \epsilon_0 r_0}\left(1-\frac{1}{n}\right)$
  
  where:
  
  
  $N_A$ = Avogadro constant;
  M  = Madelung constant
  $z^+$ = charge number of cation
  $z^−$ = charge number of anion
  e = charge on an electron (elementary charge)
  $\epsilon_0$ = permittivity of free space
  $r_0$ = distance to closest ion
  n  = Born exponent, typically a number between 5 and 12, determined experimentally or derived theoretically.
  


(Note that I've inverted the formula from Wikipedia's version to match your reaction.)

None of these can give you a negative lattice enthalpy.

How could it happen? Well, you'd need something where the solid is less stable than the gas-phase ions. Consider that in an ionic solid, the electrostatic attraction between anion and cation is always favorable. Since gas-phase ions aren't stable by themselves, I don't see how this is possible.