How can ionic bonding in lanthanides occur without valence orbitals available for overlap?
Lanthanide atoms also have $5d$ and $6s$ valence orbitals available for bonding, and these can overlap effectively with ligands.  Whether this overlap leads to a bonding orbital encompassing both metal and ligand (covalent bond) or transfer if the electrons to a much low-energy ligand orbital (ionic bond), a bond is formed either way.  The bond may not be entirely ionic but is entirely (or nearly so) derived from the $5d$ and $6s$ orbitals of the lanthanide atom instead of the $4f$ orbitals.
Lanthanides do tend to form complexes with high coordination numbers, exceeding the coordination number of six commonly seen with $d$-block transition metal complexes.  From here:

Lanthanides share many similar characteristics. These characteristics include the following:
...
Adoption of coordination numbers greater than 6 (usually 8-9) in compounds
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A preference for more electronegative elements (such as O or F) binding
...

The coordination number thereby exceeds the number of separate covalent bonds that can be formed from $5d$ and $6s$ valence orbitals alone, so a limited number of covalent bonds must be de-localized over the many linkages.  This would lead to the bonds defaulting towards a more predominantly ionic mode, requiring the ligand orbitals to be low-energy and associated with highly electronegative ligand atoms (also noted in the above excerpt).