Yes, the gravitational tidal tensor $T_{i j}$ (or $T_j^i$ with appropriate lowering of the index) is symmetric.
$$ T_{i j}=-\frac{\partial^2 \phi}{\partial x^i \partial x^j} $$
Since the gravitational potential $\phi$ is a continuous, smooth function in regions of space free of mass, the order of differentiation does not matter (Clairaut's Theorem or Schwarz's Theorem):
$$ \frac{\partial^2 \phi}{\partial x^i \partial x^j}=\frac{\partial^2 \phi}{\partial x^j \partial x^i} $$
Thus, $T_{i j}=T_{j i}$.