The point here is that definable real numbers are definable using the entire strength of might of the set theoretic universe; whereas computable real numbers are only allowed to access the natural numbers and their very very very rudimentary properties (since computable functions are only $\Sigma_1$ definable functions over $\Bbb N$).
Let $\varphi_n$ enumerate the sentences in the language of arithmetic. Now consider the real number whose $n$-th digit in the decimal expansion is $1$ if and only if $\Bbb N\models\varphi_n$, and $0$ otherwise. So it is a number in $[0,1]$.
This number is of course definable in the language of set theory, since the set of true sentences in $\Bbb N$ is definable; but it is not a computable real number since there is no computable function telling us what is true in $\Bbb N$ and what isn't (not even arithmetical, to be more accurate).
We can also take the following approach, as I suggest in the comments to the original question.
Note that every computable real lies in $L$, by absoluteness arguments (every computable functions lies there), and in $L$ there is a definable well-ordering of the reals (even with a $\Delta^1_3$ definition!), so there is a least real in the canonical well-ordering which is not definable.
Since the set "the real numbers which also lie in $L$" cannot change between models of $\sf ZF$ with the same ordinals, this set always has a canonical, definable well-ordering in any model of $\sf ZF$, and this indeed gives us a definition of a real number which is non-computable.
You can also argue that various generic reals are non-computable but definable, if you're willing to go this far as to consider different set theoretic universes (or at least one which can be seen as a nontrivial generic extension of some inner model).
For example Jensen reals are definable (they are the unique solution to a $\Pi^1_2$ predicate) but not computable.
Similarly, you can consider the iterated forcing that at the $n$-th step does the lottery sum between forcing $2^{\aleph_n}=\aleph_{n+1}$ and forcing $2^{\aleph_n}=\aleph_{n+2}$, at the limit step take a finite support limit, and consider the real number whose $n$-th decimal digit $1$ if and only if $2^{\aleph_n}=\aleph_{n+1}$, and $0$ otherwise.
This is a Cohen real which is definable, since it encodes the continuum below $\aleph_\omega$; but of course it is not computable by genericity arguments.
Note that this gives a very peculiar example of a real number, the one encoding the continuum function below $\aleph_\omega$. It is always definable, but in different models of $\sf ZFC$ it wil have different values, sometimes they will be computable (e.g. if $\sf GCH$ holds) and sometimes they could be non-computable (as above).
So this gives us a definition of a real number which is not provably computable and not provably uncomputable!
Chaitin's number is uncomputable, therefore transcendental. From the viewpoint of number systems, all transcendental numbers are pretty much the same.
Assuming $c$ is the base, you're looking at the numbers of the form
$$a_n c^n + \dots + a_1 c + a_0 + a_{-1} c^{-1} + \dots + a_{-k} c^{-k}$$
where all digits $a_i$ are rationals (you could limit that further to integers, or just integers in a fixed range).
Because $c$ is transcendental, two numbers of this form are equal if and only if all of their coefficients are equal. For example, the only way to write $5$ is $a_0 = 5$ and all other $a_i$ being 0. If you limit $a_i$ to a range, say $\{0,1,2,3\}$, then the set will not be closed under addition $2 + 3$.
Assuming you allow any rational number (or any integer number) you can add and multiply numbers of this form, pretty much the same way you operate on polynomials. They are a ring isomorphic to Laurent polynomials.
This set is not closed under division. To be able to divide, you can consider expressions of the form $f(c)/g(c)$ where $f,g$ are polynomials with rational coefficients. The arithmetic on those expressions will follow the same laws as arithmetic on rational functions; the field $\mathbb Q(c)$ is isomorphic to the set of rational functions $\mathbb Q(x)$.
This is the reason why we don't use transcendental numbers as a base - it's really studying polynomials and rational functions.
All of the above applies to any transcendental number, e.g. you could use $c=\pi$ instead of $c=\text{Chaitin's constant}$. Unlike $\pi$, Chaitin's number is noncomputable. It is transcendental over computable numbers, and not just rational numbers. Therefore, for Chaitin's constant you could make $a_i$ computable, and not just rational, and the same reasoning would apply.
Best Answer
First of all, there's a crucial vagueness here: what does "definable" mean? For example, in a precise sense it is consistent with $\mathsf{ZFC}$ that every real number is definable, the saving grace being that the relevant sense of "definable" is (per Tarski) too powerful to be subject to the obvious counting argument.
But as interesting as it is, this isn't really the core problem with your question. Let's fix some "reasonably rich" notion of definability; for example, the following will (I suspect) satisfy all the basic requirements one might have:
I'm using $\omega_{17}$ here just as some "really big" ordinal. Feel free to go bigger or smaller as you prefer, and see this old post of mine if you're unfamiliar with the notion of definability in a structure. Note that all computable reals are indeed definable in this sense (this is basically just a tweak to Post's theorem).
Let $\mathcal{D}$ be the set of definable (in the above sense) reals and $\mathcal{C}$ be the set of computable reals. Now here's the truly serious problem:
To compare countable sets in a meaningful way, we usually need to turn to some kind of asymptotic analysis. For example, the following is an almost completely precise question:
The "obvious" conjecture is that $\lambda=0$. But note that this is highly dependent on the way we choose to enumerate the formulas of set theory (and this explains the "almost" five sentences prior): we can in fact get any value we want for $\lambda$ by enumerating formulas in a perhaps pathological manner. So the question above only becomes precise once we fix an enumeration of formulas, and moreover it only becomes interesting once we justify a particular choice of enumeration (or a class of enumerations, and show that any enumeration from that class yields the same answer).
Admittedly, in terms of heuristics I find that it's almost always a good idea to assume that $\mathcal{C}$ is a negligible subset of $\mathcal{D}$. But this is very vague. So at present I will tentatively claim the following: