This can be done if you extend the triangulation of your 3-manifold $M$ to a triangulation of a 4-manifold $W$ whose boundary is $M$. You can find the basic idea in Chapter 9 of these draft notes on my web page, and there are very explicit details for the associated Hilbert space in this arXiv preprint (with Z. Wang). (In the latter paper we use cubes instead of simplices to make computer implementation easier.) There's also a summary in the notes from a talk.
The main idea is that the Witten-Reshetikhin-Turaev TQFT can be reinterpreted as a 3+1-dimensional TQFT, and the 3+1-dimensional version is "fully extended" (goes all the way down to points). Once you have a fully extended TQFT, standard techniques allow you to construct a state sum model. In this case, the state sum you get is a modified version of the Crane-Yetter state sum. (You would get a Turaev-Viro state sum for a fully extended 2+1 dimensional TQFT.)
EDIT:
If you don't want to wade through the above sources, here's the explicit formula.
Let $C$ be a modular ribbon category.
Let $D= \sqrt{\sum_i d_i^2}$, the square root of the sum of the squares of the quantum dimensions.
Choose a handle decomposition of $W^4$.
Keep in mind is the case where the $i$-handles are thickenings of the $i$-cells
of some cell decomposition of $W$, such as the dual cell decomposition to a triangulation.
Let ${\cal H}_i$ be the set of $i$-handles.
Let ${\cal L}_2$ be the set of labelings of the 2-handles by simple objects of $C$.
For fixed $\alpha\in{\cal L}_2$, let ${\cal L}_1(\alpha)$ denote the set of labelings of the 1-handles by orthogonal
basis elements of the associated vertex spaces.
(To each 1-handle is associated a $C$-picture on the linking 2-sphere,
that is, a collection $c$ of
ribbon endpoints on the 2-sphere labeled by simple objects
$a_1\otimes \cdots\otimes a_m$ according to $\alpha$.
The "vertex space" is $\hom_C(1, a_1\otimes\cdots\otimes a_m)$ or,
more canonically, the vector space associated to $(B^3; c)$, where we think
of $B^3$ as the normal fiber to the core of the 1-handle.)
Then
$$Z(W^4) = \sum_{\alpha\in{\cal L}_2} \sum_{\beta\in{\cal L}_1(\alpha)}
\prod_{h_4\in {\cal H}_4} D
\prod_{h_3\in {\cal H}_3} D^{-1}
\prod_{h_2\in {\cal H}_2} D^{-1} \,\text{Loop}(h_2, \alpha)\quad\quad\quad\quad$$
$$\quad\quad\quad\quad\prod_{h_1\in {\cal H}_1} D \,\text{Th}(h_1, \beta)^{-1}
\prod_{h_0\in {\cal H}_0} D^{-1} \, \text{Link}(h_0, \alpha, \beta) , $$
where
- $\text{Loop}(h_2, \alpha)$ is the loop value (quantum dimension) of the simple object $\alpha(h_2)$;
- $\text{Th}(h_1, \beta)$ is the evaluation of generalized theta graph with the two vertices
labeled by $\beta(h_1)$ and $\overline{\beta(h_1)}$ (in the case of a generic cell decomposition where each 1-cell is incident to three 2-cells, this is just an ordinary theta graph); and
- $\text{Link}(h_0, \alpha, \beta)$ is the evaluation of the graph in $S^3$ corresponding to the link
of the 0-handle $h_0$, labeled according to $\alpha$ and $\beta$.
If $W$ has a boundary $M$ then we can place a labeled link $L$ in the boundary, and the labels of the link play a role similar to the labels of the 2-handles above. In this case the state sum computes $WRT(M, L)$.
For a generic cell decomposition of $W$ (i.e. dual to a triangulation), the $\text{Link}(h_0, \alpha, \beta)$ factor is an evaluation of a labeled ribbon graph which looks like the 1-skeleton of a 4-simplex. If we resolve the five 4-valent vertices oin this graph into pairs of 3-valent vertices, then this becomes the "15-j symbol" used by Crane and Yetter. (Keep in mind that Crane and Yetter use a different normalization which obscures the relation to the WRT invariant.)
On the other hand, if we choose a handle decomposition of $W$ with a single 0-handle and several 2-handles (no 1- 3- or 4-handles), then it is easy to see that the above state sum reduces to the Reshetikhin-Turaev surgery formula. The factor $\text{Link}(h_0, \alpha, \beta)$ is the generalized Jones polynomial (evaluated at a root of unity) in this case.
A similar variant yields the Turaev "shadow" state sum.
