By Löwenheim–Skolem theorem every first order theory must have a countable model.
Question: Is there a result stating that every first order theory must have a pointwise definable model? i.e. a model in which every object is definable in a parameter free manner?
If the answer is to the negative then:
Question: is there a criterion that can characterize all first order theories that defies the above feature?
EDIT: although my original intention was about definability in the language of the meta-theory in which the model for the theory in question is constructed, and I take the meta-theory to be ZFC. However, the question is understood along the standard lines to be about definability in the language of the theory in question. And answers had been put forth along that line.
Question: along the standard notions, what would make a theory defy having a definable pointwise model ? is there a common criterion that all such theories enjoy?
The answer to the stated question: "Does every (consistent) first-order theory have a pointwise definable model?" is No. Examples were given by Asaf Karagila in the comments, but I'll elaborate a bit.
The easiest example is the theory of an infinite set in the empty language (with just a symbol $=$ for equality). No model of this theory has any element which is definable without parameters. Indeed, any element of a structure $M$ which is definable without parameters is fixed by all automorphisms of $M$. But for any infinite set $A$ and any element $a\in A$, there is an automorphism of $A$ moving $a$.
Another example is the theory DLO of dense linear orders without endpoints in the language $\{\leq\}$. Again, no model has any element which is definable without parameters. Here we can note that if $\varphi(x)$ were a formula defining a unique element in some model, then since DLO is a complete theory, $\varphi(x)$ would define a unique element in every model, in particular in the unique (up to isomorphism) countable model of DLO, $(\mathbb{Q},\leq)$. But by Cantor's back-and-forth argument, for any $q,q'\in \mathbb{Q}$, there is an automorphism of $\mathbb{Q}$ moving $q$ to $q'$.
But in the comments, you specified:
This is an extremely unusual use of language. Readers of your question will naturally assume that "definable" means the standard thing: definable in the language of the first-order theory in question.
It also makes the question pretty vague, since you haven't specified the meta-theory. The answer obviously depends on which theory we're using to "construct models". e.g. if that theory can't prove the completeness theorem for first-order logic, then it can't even prove that an arbitrary consistent theory has any models, much less pointwise definable ones.
But OK. Let's assume the meta-theory is ZFC. The argument is so simple that it's easy to see that any natural set theory which is strong enough to prove the completeness theorem and downward Löwenheim-Skolem will work just as well.
If a theory $T$ in a countable language is consistent, it has a countable model. So it has a model with domain $\mathbb{N}$. Since every natural number is definable without parameters in ZFC, $T$ has a pointwise definable model in your sense.