Suppose we have a linear system of differential equations governed by a matrix $A$:
$$ \frac{d\mathbf{x}}{dt} = A\mathbf{x}, \ \mathbf{x}(t=0) = \mathbf{x}_0$$
We know that the solution is
$$ \mathbf{x}(t) = e^{At}\mathbf{x}_0. $$
Now define:
$$ D:= \frac{\|\mathbf{x(t)}\|}{\|\mathbf{x}_0\|} $$
I know that we can find an upper bound for $D$ as follows:
$$ \| \mathbf{x}(t) \| \leq \|e^{At}\| \|\mathbf{x}_0\| \leq e^{\|A\|t} \|\mathbf{x}_0\| $$ $$ \therefore D \leq e^{\|A\|t}$$.
My question is: how can we find a lower bound for $D$? In the specific case I'm considering, $A$ is upper triangular (sometimes strictly upper triangular) and non-diagonalizable.
It is rather simple to give a lower bound of the kind of your upper bound.
Indeed, since $\mathbf{x}_0=e^{-At}\mathbf{x}(t)$, similarly $$ \|\mathbf{x}_0\|\le e^{\|A\|\cdot|t|} \|\mathbf{x}(t)\|\quad\Rightarrow\quad \|\mathbf{x}(t)\|\ge e^{-\|A\|\cdot|t|}\|\mathbf{x}_0\|. $$
You should also replace $t$ by $|t|$ in your question.