$T_3(t)$ is known, and it is possible to calculate an arbitrary $n$th order derivative of $T_3(t)$.
$n_1$, $d_1$, and $d_2$ are constants, and $S_3(s)$ is the Laplace transform of $T_3(t)$.
Is there a way to find the inverse Laplace transform of the following expression in terms of $T_3(t)$ and its derivatives?
$$\frac{n_1S_3(s)}{s^2+d_1s+d_2}$$
If it is possible, how does one go about finding it?
The best you can do is use convolutions. (Note that I modified your question slightly because calling both the function and its Laplace transform by the same name is a bad habit and potentially dangerous, and technically wrong. Just changing the input variable name is not enough.)
The convolution of two functions $f$ and $g$ is: $$ (f \ast g)(t) = \int_0^t f(t - \tau) g(\tau) \, d\tau$$
It is a known fact that the Laplace transform of a convolution is equal to the product of the Laplace transforms. In other words, $$\mathcal L\{f \ast g\} = \mathcal L\{f\} \cdot \mathcal L\{g\}.$$
Let $f \ast g$ denote the convolution of $f$ and $g$, let $\mathcal L$ denote the Laplace transform operator, and let $F$ and $G$ denote the Laplace transforms of $f$ and $g$, respectively.
Then since $\mathcal L\{f \ast g\} = F(s)G(s),$ we therefore have $$\mathcal L^{-1}\{F(s)G(s)\} = (f \ast g)(t) = \left(\mathcal L^{-1}\{F\} \ast \mathcal L^{-1}\{G\}\right)(t).$$
In this case: \begin{align*} \mathcal L^{-1}\left\{\frac{n_1 S_3(s)}{s^2 + d_1s + d_2}\right\} &= \mathcal L^{-1}\left\{\frac{n_1}{s^2 + d_1s + d_2} \cdot S_3(s)\right\}\\[0.3cm] &= \left(\mathcal L^{-1}\left\{\frac{n_1}{s^2 + d_1s + d_2}\right\} \ast \mathcal L^{-1}\left\{S_3\right\} \right)(t) \end{align*} We already know that $\mathcal L^{-1}\{S_3\} = T_3(t)$. Now we just need to find $\mathcal L^{-1} \left\{\dfrac{n_1}{s^2 + d_1s + d_2}\right\}$. Start by completing the square: $$ \frac{n_1}{s^2 + d_1s + d_2} = \frac{n_1}{\left(s + \frac{d_1}2\right)^2 + d_2 - \frac{d_1^2}4} $$ From here I think it depends on the sign of $d_2 - \dfrac{d_1^2}4$. You can do some more algebraic manipulations and then use a Laplace transform table. I didn't take it all the way but it looks like you'll get something involving exponentials and sine, or exponentials and hyperbolic sine - again, depending on the sign of $d_2 - \dfrac{d_1^2}4$.