This is a good question; when $q$ is not formal, I believe that Tor-groups of $U_q(\mathfrak{g})$ can differ from $U(\mathfrak{g})$ considerably. As Mariano said, if we take the formal quantum group, with $q=e^\hbar$, the answer is the same as for $U(g)$, so I'll address non-formal $q$.

I interpret your question as asking for a free resolution of $k=\mathbb{C}$, since it is then easy to compute Tor from this. The following is a free resolution of $k$ over $U=U_q(\mathfrak{sl}_2)$.

Recall that $U$ has generators $E$,$F$, and $K$, $K$ is invertible, and relations $EF-FE= \frac{K-K^{-1}}{q-q^{-1}}$, $KE=q^2EK$, and $KF=q^{-2}FK$. Here is a resolution of $k=\mathbb{C}$ as a module over $U$ via $\epsilon$.

$\begin{align*}
M_{-1}&=k\\
M_0&= U\\
M_1 &= U\otimes E \oplus U\otimes F \oplus U\otimes (K-1))\\
M_2 &= U\otimes (E\otimes F- F\otimes E - \frac{K^{-1}+1}{q-q^{-1}}\otimes (K-1))\\
&\oplus U\otimes (q^{-2}(K-1)\otimes E-E \otimes (K-1)+(q^{-2}-1)\otimes E)\\
& \oplus U\otimes (q^2(K-1)\otimes F - F\otimes (K-1) + (q^2-1)\otimes F)\\
M_3&=U \otimes \Big[(1-K)\otimes(E\otimes F- F\otimes E - \frac{K^{-1}+1}{q-q^{-1}}\otimes (K-1))\\
&+F\otimes(q^{-2}(K-1)\otimes E-E \otimes (K-1)+(q^{-2}-1)\otimes E)\\
&-E\otimes(q^2(K-1)\otimes F - F\otimes (K-1) + (q^2-1)\otimes F))\Big]
\end{align*}$

Having written all that out, I'm depressed that it's typeset so ugly =[. I hope the notation is clear. The tensorands on the right are just formal symbols, but they suggestively tell you what the differential is: you just multiply the $U$-coefficient by first tensorand, to move from $M_k$ to $M_{k-1}$. e.g, for $X\in U$, we have $d(X\otimes E)=XE$.

It's a good exercise to check that it is a free resolution; I will omit the computations. For instance, exactness at $M_0$ is the claim that $Ker(\epsilon) = <E,F,K-1>$. Exactness at $M_1$ has to do with the relations of $U$. Note that the free rank over $U$ is the same as in the C-E complex.

To finish up, note that $k\otimes_U -$ on this complex just means applying $\epsilon$ to the $U$ term everywhere, so we get:
$$\mathbb{C}v^3_1 \to \mathbb{C}\langle v^2_1, v^2_2, v^2_3,\rangle \to\mathbb{C}\langle v^1_1,v^1_2,v^1_3\rangle \to\mathbb{C}v^0_1,$$

where I have named the basis vectors of $M_k$ $v^k_?$, in the order they appear above, e.g. $v^1_1=E$.

Okay, note that: $d(M_1)=0$, but $d(M_2)=M_1$, because $\epsilon(\frac{K^{-1}+1}{q-q^{-1}})=\frac{2}{q-q^{-1}}$, $\epsilon(q^{-2}-1)$, and $\epsilon(q^2-1)$ are non-zero. Finally, $d(M_3)=0$ (as it had better, to form a complex!)

Thus we have

$$Tor_j(k,k)=\left\{\begin{array}{ll}k,& j=0,3\\0,&j=1,2\\&0,j\geq 4\end{array}\right.$$

Note that I computed the above only for $\mathfrak{sl}_2$, but it should be possible to do the same for arbitrary quantum groups, by similarly following one's nose. I suppose, however, that one would need in that case to use the cubic $q$-serre relations, e.g. between $E_1$ and $E_2$ for $sl_3$, which are a pain.