I find an area of $.245463$ for the intersection when each $T_i$ has a vertex at distance $3.15835$ from the center of $T$. This seems to be maximal among the symmetric options.

The gray triangle in the diagram is $T$, the black triangles are the $T_i$ (with their specified far vertices outside the area of the diagram), and the blue nonagon is the desired intersection.

[![enter image description here][1]][1]

If the distance is $(1+2b)R$, where $R$ is the circumradius of $T$, we can write the area exactly as
$$4\left(2b - b^2 - \frac{3}{4b^2-1} - \frac{8b^2-4b-2}{12b^2 -1}\right)$$

which is easy to maximize numerically.

For a quicker approximate answer, we can assume that the area of intersection is a circle, tangent to the two long sides of $T_1$ at $(r,\theta)$ in polar coordinates. Then we can solve for $r,R,\theta$ satisfying
\begin{align}
3R^2 \sqrt{3} / 4 &= \,1, \text{ for circumradius of }T;\\
r(\sec \theta - 2)/3 &= R, \text{ for centroid of }T_1;\\
r^2(1+\sec \theta)^2 \cot \theta &= \,1, \text{ for area of }T_1
\end{align}
and the area and distance quoted above are $\pi r^2\simeq .225$ and $r \sec \theta \simeq 3.17$.

  [1]: https://i.sstatic.net/kLHsD.png