Description of Tor via the derived category? If $A,B$ are objects of an abelian category $\mathcal{A}$ and $n \in \mathbb{N}$, there is a very nice and useful description of $\mathrm{Ext}^n(A,B)$. Namely, it is just the set of morphisms $A \to B[n]$ in the derived category $D(\mathcal{A})$. For example, this gives more elegant formulations of the Yoneda product $\mathrm{Ext}^n(A,B) \otimes \mathrm{Ext}^m(B,C) \to \mathrm{Ext}^{n+m}(A,C)$ and Serre duality on a smooth projective scheme $X$.
Now let $\mathcal{A}$ be an abelian $\otimes$-category with enough projectives / flat objects. Is there a similar description of $\mathrm{Tor}_n(A,B)$ using the derived category? More precisely, can we manipulate $A,B$ inside of $D(\mathcal{A})$ (using shifts, Homs, tensor products etc.) to get the abelian group $\mathrm{Tor}_n(A,B)$ without talking about projective resolutions?
 A: So note that if $\mathscr{A}$ is an abelian tensor category, then $\mathrm{Tor}_n(-, -)$ is an object of $\mathscr{A}$, not an abelian group. The functors you've listed (derived tensor, shifts, homs) either land in the derived category $D(\mathscr{A})$ or in the category of abelian groups, so in the generality you've indicated there can't be a way of recovering $\mathrm{Tor}_n(-,-)$ with the operations you've listed.
This means we have two options. First option is we expand our repertoire of functors we allow. In particular, the functors we need are the cohomology functors $H^n : D(\mathscr{A}) \to \mathscr{A}$, since then we can obviously recover $\mathrm{Tor}_n(-,-)$ as the composite $H^{-n}(- \otimes^{L} -)$. But that's tautological and not satisfying. 
The second option is we put some additional hypotheses on our category so that one of the functors we've already allowed ourselves actually takes values in $\mathscr{A}$. Derived tensor and shifts will land in the derived category again, so that won't help and the only hope is if hom takes values in $\mathscr{A}$. In other words, we'd need $\mathscr{A}$ to be enriched over itself or something. I'm not quite sure what that means since we might need the abelian group enrichment to coincide with the enrichment over $\mathscr{A}$, which doesn't make sense unless objects of $\mathscr{A}$ are already abelian groups, so to avoid saying incorrect things I'm just going to go ahead and assume that $\mathscr{A}$ is the category of modules over a commutative ring $R$. Now hom's do take values in $\mathscr{A}$, so we have some hope. 
So now it would be sufficient to construct the cohomology functors $H^n$ from what we have already. One way to have that would be for $H^n$ to be representable on $D(R)$. But it is! It's not so difficult to see from definitions that $\mathrm{Hom}_{K(R)}(R[-n], A) = H^n(A)$ naturally for all complexes $A$, where $R[-n]$ is the complex which is just $R$ in degree $n$ and $K(R)$ is the homotopy category of complexes of $R$-modules. Moreover clearly $\mathrm{Hom}_{K(R)}(R[-n],-) = \mathrm{Hom}_{D(R)}(R[-n], -)$, so we see that $R[-n]$ represents (or, maybe to be more terminologically correct, corepresents) the functor $H^n$ on the derived category $D(R)$ of $R$-modules. 
So, putting it all together, we get
$\mathrm{Tor}_n(-, -) = \mathrm{Hom}_{D(R)}(R[n], - \otimes^L -)$.
That's still not quite as satisfying as the description of $\mathrm{Ext}$ so I'm not sure if that's the kind of thing you were looking for. 
