If energy is quantized, does that mean that there is a largest-possible wavelength?
since energy is quantized
You have a misunderstanding here on what quantization means. At present in our theoretical models of particle interactions all the variables are continuous, both space-time and energy momentum. This means they can take any value from the field of real numbers. It is the specific solution of quantum mechanical equations, with given boundary conditions that generates quantization of energy.
The same is true for classical differential equations, as far as frequencies go. Sound frequency can take any value, and its quantization in specific modes depends on the specific problem and its boundary conditions.
There exist limits given by the value of the constants that are used in elementary particle quantum mechanical equations. It is the Planck length and the Planck time
the reciprocal of the Planck time can be interpreted as an upper bound on the frequency of a wave. This follows from the interpretation of the Planck length as a minimal length, and hence a lower bound on the wavelength.
which are at the limits of what we can see in experiments and study in astrophysical observations, but these are another story.
You have to keep in mind that the relation
$$ E = hf $$
holds only true for photons. Photons - generally - can have arbitrary energies, so they can have arbitrary frequencies as well.
When you think of quantization, you think of quantization of an observable for a specific system. An one-dimensional harmonic oscillator for example has the quantized energies
$$ E_\textrm{Osz} = h f_\textrm{Osz} \left(\frac{1}{2} +n\right) $$
where $n$ is $\mathbb Z_{0}^{+}$, and $f_{Osz}$ is the frequency of the oscillator! So the energy has discrete values it can hold, but remember, we are talking about the energy of the oscillator, not of a photon. If you now ask the question: What is the minimum energy (and therefore maximum wavelength) for a photon to get absorbed by the oscillator, the answer would be:
$$ E_\textrm{PhMin} = 1\cdot h f_\textrm{Osz} \ \rightarrow \ \lambda_\textrm{PhMax} = c/f_\textrm{Osz} $$
because that is the difference between two niveaus of the oscillator.
If you look at another system your answer will be different. How observables (like energy) are quantized is dependent on the system.
tl'dr:
There is no maximum wavelength for photons.
Edit: At least not because of the Planck Relation. If there is a maximum wavelength for photons, the reasons for it will have nothing to do with the implications of your question. I could be that - at sufficiently small energies - photons show non-trivial effects which restricts another loss of energy.
(Please note that I used simplified assumptions, for instance we are in vacuum etc.)