This is a partial answer to this question:

I was always under the impression that a type-level pair must depend on having some kind of well ordered infinite class of objects, well at least this was the experience I found with the Quine-Rosser pair, and in addition it should presuppose Extensionality (like Quine-Rosser pairs, or even Holmes's $1$-high type pairs). However, it's only today when I came to realize that this is not necessarily correct. 

I think I can cook up a pair that fulfills the aforementioned criteria.

Lets work in $\sf ZFA \neg C$.

Now, let $F$ be a total injective function that sends sets to nonempty sets of nonempty sets not having sets $1,2$ among their elements. That is: $$ F(x)=y \to \\ \exists z\, (z \in y) \land \forall z (z \in y \to \exists u \, (u \in z) \land 1 \not \in z \land 2 \not \in z)$$ 
Now, we define an "inserter" function $I_\alpha$ that inserts $\alpha$ to all elements of a set, that is:

$$I_\alpha(x)=\{y \cup \{\alpha\} \mid y \in x \}$$

Now, we define the following pair:

$$(x,y) = I_1 (F(x)) \cup I_2 (F(y))$$

We can easily retrieve the $\alpha^{th}$ projection of $(x,y)$: We take the set of all elements of $(x,y)$ that have $\alpha$ among their elements, apply the de-inserter function $I_\alpha^{-1}$ on it, then apply $F^{-1}$ and we get the $\alpha^{th}$ projection of $(x,y)$.

We can extend that to any $\lambda$-tuple: $$(x_1,x_2,...)^\lambda = \underset {\alpha < \lambda} \bigcup I_\alpha(F(x_\alpha))$$

Of course, to answer the above question, we can simply take the tuple to be: $$\{\{(x_1,x_2,...)^\lambda\}\}$$
The main drawback is that this definition is not that general, for instance it doesn't work in the usual known models of $\sf NFU$, where we have strictly more empty objects than nonempty. In such models I only know of pairs with height of at least $2$ that can do the job.