Yes. No problem. This is the $q$-analogue of the symmetriser. In terms of $T_i$ we have $$ \sum_{\pi\in frac{1}{[n]!}\sum_{\pi\in S_n} q^{\ell(\pi)} T_\pi$$ where for $T_\pi$ we take a reduced word for $\pi$.\pi$ and $\ell(\pi)$ is the length of a reduced word.
There is another presentation for the Hecke algebra which I am used to writing as generators $u_i$ and defining relations $$u_i=[2]u_i$$ $$u_iu_j=u_ju_i\qquad\text{for $|i-j|>1$}$$ $$u_iu_{i+1}u_i-u_i=u_{i+1}u_iu_{i+1}-u_{i+1}$$
Strictly speaking this is the subring of the Hecke ring which is invariant under the bar involution. This algebra is defined over $\mathbb{Z}[\delta]$. The Hecke algebra is the algebra over $\mathrm{Z}[q,q^{-1}]$ obtained by the specialisation $\delta\mapsto q+q^{-1}$.
Then we have $u_i=C_i$ or maybe $u_i=-C_i$.
The Temperley-Lieb algebra is the quotient by $u_iu_{i\pm 1}u_i=u_i$.
This has an involution given by $u_i\leftrightarrow \delta-u_i$.
To define the idempotents we need to divide by $[n]!$.
Define $R_i(k)=1-\frac{[k]}{[k+1]}u_i$. Then these satisfy the Yang-Baxter equation $$R_i(r)R_{i+1}(r+s)R_i(s)=R_{i+1}(s)R_i(r+s)R_{i+1}(r)$$
Using this you can write everything (until someone tells me otherwisewell a lot, at least) explicitly. For example the idempotents can be written
$$1\qquad R_1(1)\qquad R_1(1)R_2(2)R_1(1)\qquad R_1(1)R_2(2)R_1(1)R_3(3)R_2(2)R_1(1)$$ and so on. You get another sequence of idempotents by applying the involution. Under the involution we have $R_i(k)\leftrightarrow R_i(-k)$. The three string idempotent is $$\left(\frac{u_1}{\delta}\right)\left(1-\delta u_2\right)\left(\frac{u_1}{\delta}\right)$$ which gives the relation for the Temperley-Lieb algebra.
