Based on my knowledge, lattice energy is proportional to the multiplcation of the charge of the ions, divided by the sum of the radius of ions, as follows.
Since iodide has a larger radius than fluoride, by this equation lithium iodide would have a less exothermic lattice energy (lower lattice energy).
However, I also learned that an ionic bond with higher covalent character is stronger (i.e. more exothermic lattice energy). Since the electron cloud of iodide ion is larger and is more polarisable, lithium iodide should have greater covalent character than lithium fluoride and thus its lattice energy should be more exothermic (higher lattice energy)