I would like to know which motivic cohomology groups of complex numbers are nonzero and ("better") nontorsion, i.e., for which $(i,j)$ the $i$th cohomology of the complex ${\mathbb{Q}}(j)$ over $\mathbb{C}$ is not zero. I would like to know both which of these groups are known to be nonzero (and also known to be "large") unconditionally and also which of them shouldn't vanish according to some conjectures. Certainly, my question can be easily translated into the language of $K$theory.
This is going to be a slightly extended explanation, I apologize. The short version is basically that little is actually known (and even that is hard to prove), but conjecturally everything permitted by BeilinsonSoulé vanishing should be infinitedimensional.
What is known unconditionally (I bet everybody knows the first three items and I make no claim that this is an exhaustive list):
the groups ${\rm H}^i({\rm Spec} \mathbb{C},\mathbb{Q}(n))$ vanish for $i>n$ (follows from Nisnevich cohomology dimension reasons).
the groups ${\rm H}^n({\rm Spec} \mathbb{C},\mathbb{Q}(n))$ are uniquely divisible groups, uncountablydimensional as $\mathbb{Q}$vector spaces. (Milnor Ktheory, related to elements in $\mathbb{C}^\times$)
the groups ${\rm H}^{1}({\rm Spec} \mathbb{C},\mathbb{Q}(n))$ (corresponding to weight $n$ part of $K_{2n1}$) are nontrivial infinitedimensional $\mathbb{Q}$vector spaces. The Ktheory of $\mathbb{C}$ contains the Ktheory of $\overline{\mathbb{Q}}$ as direct summand, and the groups ${\rm H}^{1}({\rm Spec}\overline{\mathbb{Q}},\mathbb{Q}(n))$ are vector spaces of countable dimension, detected by Borel's regulator (this computation is due to Borel). The rigidity conjecture voices the expectation that the inclusion $\overline{\mathbb{Q}}\subset \overline{\mathbb{C}}$ induces an isomorphism on ${\rm H}^{1}(,\mathbb{Q}(n))$ for $n\geq 2$.
C. Soulé. $p$adic Ktheory of elliptic curves. Duke Math. J. 54 (1987), 249269. His Theorem 2.1 provides unconditional lower bounds for highdimensional Kgroups (related to the argument given below). From curves over number fields we would get contributions in ${\rm H}^2({\rm Spec}\mathbb{C},\mathbb{Q}(n))$, which is just above the Borel case above. For example, Soule's unconditional result tells us that weight 82 in $K_{162}(\mathbb{C})$ is unbounded (as well as all the bigger weights in the appropriate Kgroups...).
What is expected: (I think)
Of course, the first thing to mention is that the BeilinsonSoulé vanishing conjecture predicts the vanishing of ${\rm H}^i({\rm Spec}\mathbb{C},\mathbb{Q}(n))$ for any $i\leq 0$ except for $i=n=0$. In particular, it would be conjectured that the motivic cohomology of $\mathbb{C}$ can only be nontrivial for $1\leq i\leq n$.
Now I would like to claim that whatever nontrivial motivic cohomology groups are allowed by BeilinsonSoulé vanishing also contain nontrivial elements and are in fact infinitedimensional vector spaces, depending on conjectures of Beilinson and Serre.
First step: if $X/F$ is a smooth projective variety over a number field $F$, then the function field $F(X)$ is finitely generated and therefore embeds into $\mathbb{C}$.
Second, the rational motivic cohomology of a function field embeds into the rational cohomology of $\mathbb{C}$. This follows from the existence of transfers on motivic cohomology; in a finite field extension, composition of the transfer with the natural presheaf restriction is multiplication by the degree and therefore an isomorphism for rational coefficients. So we cannot kill any motivic cohomology by finite extensions. On the other hand, finite transcendental extensions of algebraically closed fields also induce split injections (by evaluating on a rational point of the respective affine space). So, no rational motivic cohomology is lost by passing from the function field to $\mathbb{C}$.
Third, (in the range we are interested in) the motivic cohomology of a smooth projective variety $X/F$ embeds into the motivic cohomology of its function field. This is essentially a form of the Gersten resolution. For ease of citation, I would replace motivic cohomology by Nisnevich cohomology of the Milnor Ksheaf (see my answer to this MOquestion: Motivic cohomology and cohomology of Milnor Ktheory sheaf) and use the Gersten resolution of the latter.
Fourth, we now employ the conjectures of Beilinson and Serre (but note that all reduction steps so far were unconditional). For more information, see Schneider's survey "Introduction to the Beilinson conjectures" (see here) or this great MOanswer: Beilinson conjectures. If we have a smooth projective variety $X/F$, Beilinson's conjecture predicts an isomorphism between "the integral part" of motivic cohomology of $X$ with DeligneBeilinson cohomology of $X(\mathbb{R})$ (under suitable condition on the indices). Put differently, the rank of the integral part of motivic cohomology (a subspace in ${\rm H}^{i+1}(X,\mathbb{Q}(n))$) should be equal to the order of vanishing of the Lfunction at $s=i+1n$, assuming that $i/2+1<n$. Note that this condition is satisfied for our range of interest $i\geq 1$ because $i/2+1<i+1\leq n$. Now in Schneider's survey paper we also find (p.6) that  assuming conjectures of Serre on the Lfunctions of varieties over number fields  this order of vanishing is given by $\dim {\rm H}^i(X(\mathbb{C}),\mathbb{C})^{(1)^{i+1n}}$. For odd $i$, we can replace this by $1/2\dim {\rm H}^i(X(\mathbb{C}),\mathbb{C})$.
Summing up, the conjectures of Beilinson and Serre imply that the motivic cohomology group ${\rm H}^{i+1}({\rm Spec}\mathbb{C},\mathbb{Q}(n))$ is at least as big as $\dim {\rm H}^i(X(\mathbb{C}),\mathbb{C})^{(1)^{i+1n}}$ (resp. $1/2\dim{\rm H}^i(X(\mathbb{C}),\mathbb{C})$ for odd $i$) for any smooth projective variety $X$ over a number field. Now we can just take abelian varieties over number fields: the dimensions of the cohomology in a fixed degree $i$ (and then also the $\pm 1$ eigenspaces for even $i$) grow like a binomial coefficient times dimension of the abelian variety times the degree of the number field. So, we can find abelian varieties where (conjecturally, Serre) the order of vanishing of the Lfunction is arbitrarily big and consequently (conjecturally, Beilinson) the dimension of the motivic cohomology group ${\rm H}^{i,n}$ is arbitrarily big. (Actually, for highenough dimension, we only need the conjecture of Serre and can replace the application of Beilinson's conjecture by Soule's result mentioned above.) Since all these motivic cohomology groups of abelian varieties embed into the motivic cohomology of $\mathbb{C}$, we get an infinitedimensional vector space.
So the motivic cohomology of the complex numbers should be big because it incorporates motivic cohomology of all arithmetic schemes. At this point, though, I don't know if the conjectures would make a prediction as to countable vs uncountable dimension. (Of course, that rational cohomology has to be so big if it's nonvanishing is sort of clear: if we had finitedimensional vector spaces, there would have to be residual torsion but this can't exist by Suslin's rigidity.)
As a final remark, there is another possible conjectural way to deduce statements about the motivic cohomology of $\mathbb{C}$. In the papers of Goncharov (in particular the trilogarithm paper in the Motives Proceedings), there is a conjectural description of the Ktheory of fields. The motivation for this is that Ktheory of a field $F$ should be the cohomology of the motivic Lie algebra associated to the category of mixed Tate motives over the field (again requires BeilinsonSoulé conjecture). The conjecture says that the motivic Lie algebra should be as simple as possible: the abelianization is related to $F^\times$, and the commutator subalgebra is a free Lie algebra related to higher Bloch groups. This gives rise to explicit complexes conjecturally computing rational Ktheory of $F$ in symbollike terms. However, I am unaware that anyone has used this to produce large subspaces in the Ktheory of $\mathbb{C}$. Goncharov has used the presentation to produce elements in Ktheory of elliptic curves.
Here is some more information on the rigidity question, as requested. Explicitly, this is the question if for any field $F$ with field of constants $F_0$ (i.e. algebraic closure of prime field in $F$) the inclusion $F_0\to F$ induces an isomorphism ${\rm H}^1({\rm Spec}F,\mathbb{Q}(n))\to {\rm H}^1({\rm Spec}F_0,\mathbb{Q}(n))$ for any $n\geq 2$.
For $n=2$, this is a question about indecomposable $K_3$. In this case, we can express the regulator in terms of the dilogarithm. If we knew injectivity of the regulator (conjectures of Milnor and Ramakrishnan), then rigidity would follow from the rigidity of the dilogarithm, cf. Sections 6 and 7 of Bloch's Irvine notes on Ktheory of elliptic curves. In Sah's paper "Homology of Lie groups made discrete III", JPAA 56 (1989), pp. 269312, Conjecture 4.6, the rigidity conjecture for $K_3$ is attributed to Suslin. It appeared in scissors congruence literature because rigidity would complete the scissors congruence classification of 3dimensional hyperbolic polytopes.
In Goncharov's paper in the Motives proceedings, the rigidity is formulated as Conjecture 1.7 and attributed to Beilinson. Indeed, it's a consequence of the Beilinson conjectures as follows. Any class in ${\rm H}^1({\rm Spec}\mathbb{C},\mathbb{Q}(n))$ comes from a smooth projective variety $X/F$ over a number field $F$. The map induced on motivic cohomology is injective, so it suffices to compute dimensions. If we assume the vanishing conjecture, then the Gersten resolution also tells us that we get an isomorphism ${\rm H}^1({\rm Spec} F(X),\mathbb{Q}(n))\cong {\rm H}^1(X,\mathbb{Q}(n))$. (There is a slight gap here; actually, I want to put the integral subspace there, but that requires a slightly subtler argument.) Again, if we assume the conjectures of Beilinson and Serre, we can determine the dimension of the vector space ${\rm H}^1(X,\mathbb{Q}(n))_{\mathbb{Z}}$ as $\dim {\rm H}^0(X(\mathbb{C}),\mathbb{C})^{(1)^{1n}}$. This is exactly the same dimension as for the vector space ${\rm H}^1({\rm Spec} F_0,\mathbb{Q}(n))$ (in some sense, ${\rm Spec}F_0$ is the scheme of connected components of $X$) hence we get an isomorphism in motivic cohomology.

$\begingroup$ Thank you very much! I am also interested in rigidity of the sort you have mentioned; so, could you say more on this (or give some references)? $\endgroup$ – Mikhail Bondarko Aug 23 '17 at 15:57

1$\begingroup$ @MikhailBondarko: I added some references on rigidity and a sketch of argument how it also follows from Beilinson's conjectures. Hope this is useful. I also fixed an inaccuracy in what I wrote on the vanishing orders of Lfunctions. $\endgroup$ – Matthias Wendt Aug 24 '17 at 10:58