One should be careful with the definitions here. Notation: Given measurable spaces $(X, \mathcal{B}_X), (Y, \mathcal{B}_Y)$, a measurable map $f : X \to Y$ is one such that $f^{-1}(A) \in \mathcal{B}_X$ for $A \in \mathcal{B}_Y$. To be explicit, I'll say $f$ is $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable.
Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$, so the Lebesgue $\sigma$-algebra $\mathcal{L}$ is its completion with respect to Lebesgue measure $m$. Then for functions $f : \mathbb{R} \to \mathbb{R}$, "Borel measurable" means $(\mathcal{B}, \mathcal{B})$-measurable. "Lebesgue measurable" means $(\mathcal{L},\mathcal{B})$ measurable; note the asymmetry! Already this notion has some defects; for instance, if $f,g$ are Lebesgue measurable, $f \circ g$ need not be, even if $g$ is continuous. (See Exercise 2.9 in Folland's Real Analysis.)
$(\mathcal{L}, \mathcal{L})$-measurable functions are not so useful; for instance, a continuous function need not be $(\mathcal{L}, \mathcal{L})$-measurable. (The $g$ from the aforementioned exercise is an example.) $(\mathcal{B}, \mathcal{L})$ is even worse.
Given a probability space $(\Omega, \mathcal{F},P)$, our random variables are $(\mathcal{F}, \mathcal{B})$-measurable functions $X : \Omega \to \mathbb{R}$. The Lebesgue $\sigma$-algebra $\mathcal{L}$ does not appear. As mentioned, it would not be useful to consider $(\mathcal{F}, \mathcal{L})$-measurable functions; there simply may not be enough good ones, and they may not be preserved by composition with continuous functions. Anyway, the right analogue of "Lebesgue measurable" would be to use the completion of $\mathcal{F}$ with respect to $P$, and this is commonly done. Indeed, many theorems assume a priori that $\mathcal{F}$ is complete.
Note that, for similar reasons as above, we should expect $f(X)$ to be another random variable when $f$ is Borel measurable, but not when $f$ is Lebesgue measurable. Using $(\mathcal{F}, \mathcal{L})$ in our definition of "random variable" would not avoid this, either.
The moral is this: To get as many $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable functions $f : X \to Y$ as possible, one wants $\mathcal{B}_X$ to be as large as possible, so it makes sense to use a complete $\sigma$-algebra there. (You already know some of the nice properties of this, e.g. an a.e. limit of measurable functions is measurable.) But one wants $\mathcal{B}_Y$ to be as small as possible. When $Y$ is a topological space, we usually want to be able to compose $f$ with continuous functions $g : Y \to Y$, so $\mathcal{B}_Y$ had better contain the open sets (and hence the Borel $\sigma$-algebra), but we should stop there.
No, it is not possible. It is consistent with ZF without choice that
the reals are the countable union of countable sets. (*)
From this it follows that all sets of reals are Borel. Of course, the "axiom" (*) makes it impossible to do any analysis. As soon as one allows the bit of choice that it is typically used to set up classical analysis as one is used to (mostly countable choice, but DC seems needed for Radon-Nikodym), one can implement the arguments needed to show
(**) The usual hierarchy of Borel sets (obtained by first taking open sets, then complements, then countable unions of these, then complements, etc) does not terminate before stage $\omega_1$ (this is a kind of diagonal argument).
Logicians call the sets obtained this way $\Delta^1_1$. They are in general a subcollection of the Borel sets. To show that they are all the Borel sets requires a bit of choice (One needs that $\omega_1$ is regular).
There is actually a nice result of Suslin relevant here. He proved that the Borel sets are precisely the $\Delta^1_1$ sets: These are the sets that are simultaneously the continuous image of a Borel set ($\Sigma^1_1$ sets), and the complement of such a set ($\Pi^1_1$ sets).
That there are $\Pi^1_1$ sets that are not $\Delta^1_1$ (and therefore, via a bit of choice, not Borel) is again a result of Suslin. He also showed that any $\Sigma^1_1$ set is either countable, or contains a copy of Cantor's set and therefore has the same size as the reals. His example of a $\Sigma^1_1$ not $\Delta^1_1$ set uses logic (a bit of effective descriptive set theory), and nowadays is more common to use the example of the $\Pi^1_1$ set WO mentioned by Joel, which is not $\Delta^1_1$ by what logicians call a boundedness argument.
A nice reference for some of these issues is the book Mansfield-Weitkamp, Recursive Aspects of Descriptive Set Theory, Oxford University Press, Oxford (1985).
Best Answer
The answer below has been edited in light of other answers and comments.
There are all sorts of models of $ZFC$ in which every set is definable without parameters, including nonmeasurable sets; indeed a recent paper of Hamkins, Linetsky, and Reitz is devoted to such "pointwise definable" models.
Also, as pointed out in Theo Buehler's comment to the question, there certainly exist definable subsets of reals that are $ZFC$-provably not Borel.
As pointed out in Demer's answer, Krivine [without an inaccessible] provided a model of $ZFC$ in which every ordinal definable subset of reals is measurable. Moreover, as shown by Harvey Friedman, here, there is a model of $ZFC$ [which is a generic extension of Solovay's model] in which the following property holds:
Note that (*) implies that no non-measurable subset of reals in definable, since if $X$ is any definable subset of reals that is not measurable, then the equivalence class $\[X\]$ of $X$ modulo null sets satisfies the following two properties:
(1) $\[X\]$ definable,
(2) No member of $\[X\]$ is measurable.