Observe that
$$ \lambda_{\operatorname{max}}(e^{-A t}) = e^{-\gamma(A) t} \leq \lVert e^{-A t} \rVert \leq \alpha e^{-\lambda t} $$
where $\lambda_{\operatorname{max}}(X) = \max\{|\lambda| : \lambda \in \lambda(X)\}$ and $\gamma(X) := \min \{ \operatorname{Re}(\lambda): \lambda \in \lambda(X)\}$.
Since $A$ is lower triangular and real, it has real eigenvalues. Also since $-A$ is Hurwitz, it has real positive eigenvalues. Therefore, $\lambda$ in the exponent can be chosen as the minimum eigenvalue of $A$.
Also, if $A$ is diagonalizable, $\alpha$ can be chosen as $\alpha = \kappa(T) := \lVert T \rVert \lVert T^{-1} \rVert$ using the norm inequalities where $T$ is the matrix such that $T^{-1} A T$ is diagonal.
But, you can't change the exponent by selecting the constant in general. Because the exponential term is, well, exponential. So it will eventually catch the constant, independent of how big it is.
Edit: For any matrix $X \in \mathbb{C}^{n \times n}$, $\lambda_{\operatorname{max}}(X) = \rho(X) \leq \lVert X \rVert$. $\rho(X)$ is called the spectral radius of $X$. Also, if $\lambda$ is an eigenvalue of $A$, then $e^{-\lambda t}$ is an eigenvalue of $e^{-At}$ with the same eigenvector (Spectral Mapping Theorem).
Since $A$ is lower diagonal, $e^{-At}$ is also lower diagonal, because all powers of $A$ are lower diagonal. The diagonal elements of a lower diagonal matrix are also its eigenvalues, this is why $-A$ has all positive real eigenvalues.
Lastly, if $A$ is diagonalizable with $T$, then
$$ \lVert e^{-At} \rVert \leq \lVert T e^{-\Lambda t} T^{-1} \rVert \leq \lVert T \rVert \lVert e^{-\Lambda t} \rVert \lVert T^{-1} \rVert = \kappa(T) e^{-\gamma(A) t} $$
where $\Lambda$ is the diagonal matrix of eigenvalues of $A$. The last equation follows because any induced norm of a diagonal matrix is equal to its spectral radius.
