Is there alway exist a Lagrangian in the form of $L=T-C(\{q_i\},\{\dot q_i\},t)$, where $C$ contains no algebraical rewrite of $T$? The Lagrangian equation is
$$\frac{d}{{dt}}\frac{{\partial L}}{{\partial {{\dot q}_j}}} - \frac{{\partial L}}{{\partial {q_j}}} = {Q_j}$$
Following D'Alembert's Principle
$$\left( {\frac{d}{{dt}}\frac{{\partial T}}{{\partial {{\dot q}_j}}} - \frac{{\partial T}}{{\partial {q_j}}}} \right) - {Q_j} = 0$$
When $Q_j$ fits one of the following two conditons the Lagrangian is in the form $L=T-U$, with $U$ being the potential
$$
{Q_j} = - \frac{{\partial U}}{{\partial {q_j}}}\\
\ \\
\text{or}\\
\ \\
{Q_j} = - \frac{{\partial U}}{{\partial {q_j}}} + \frac{d}{{dt}}\frac{{\partial U}}{{\partial {{\dot q}_j}}}
$$

Suppose now I manually define a scalar term $C(\{q_i\},\{\dot q_i\},t)$ such that
$$ {\frac{d}{{dt}}\frac{{\partial (T - C)}}{{\partial {{\dot q}_j}}} - \frac{{\partial (T - C)}}{{\partial {q_j}}}} = {Q_j} - \frac{d}{{dt}}\frac{{\partial C}}{{\partial {{\dot q}_j}}} + \frac{{\partial C}}{{\partial {q_j}}}$$
and rewrite the above equation into the following form:
$$ {\frac{d}{{dt}}\frac{{\partial (T - C)}}{{\partial {{\dot q}_j}}} - \frac{{\partial (T - C)}}{{\partial {q_j}}}} = Q_j'$$
I can again obtain an expression similar to the lagrangian equation

Now can I choose this $(T-C)$ term as my Lagrangian of the system and treat $Q_j'$ as some kind of friction?
Does this mean: there alway exists a Lagrangian in the form of $L=T-C(\{q_i\},\{\dot q_i\},t)$, where $C$ contains no algebraical rewrite of $T$?
(By not an algebraical rewrite I meant $C \ne aT^b+k$)
 A: Yes, you can. But it doesn't matter if the RHS still has a $Q_j'$. You've just swapped your ignorance of $Q_j$ with your ignorance of $Q_j'$.

When $Q_j$ fits one of the following two conditons the Lagrangian is in the form $L=T-U$
$$
{Q_j} = - \frac{{\partial U}}{{\partial {q_j}}}\\
\ \\
\text{or}\\
\ \\
{Q_j} = - \frac{{\partial U}}{{\partial {q_j}}} + \frac{d}{{dt}}\frac{{\partial U}}{{\partial {{\dot q}_j}}}
$$

Writing $L=T-V$ isn't the point here. The point is : For the above two cases you mentioned, You can write $L=T-V$ satisfying:
\begin{equation}
\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0 \quad (\star)
\end{equation}
The $=0$ on the RHS is why these two cases are so special. In such cases, it's easier to obtain the equations of motion than if there were a $Q_j$ sitting on the RHS.
You can write $L=T-C$, as you did:

Suppose now I manually define a scalar term $C(\{q_i\},\{\dot q_i\},t)$ such that
$$ {\frac{d}{{dt}}\frac{{\partial (T - C)}}{{\partial {{\dot q}_j}}} - \frac{{\partial (T - C)}}{{\partial {q_j}}}} = {Q_j} - \frac{d}{{dt}}\frac{{\partial C}}{{\partial {{\dot q}_j}}} + \frac{{\partial C}}{{\partial {q_j}}}$$
and rewrite the above equation into the following form:
$$ {\frac{d}{{dt}}\frac{{\partial (T - C)}}{{\partial {{\dot q}_j}}} - \frac{{\partial (T - C)}}{{\partial {q_j}}}} = Q_j'$$

but there will still be a $Q_j'$ on the RHS. This $L$ isn't even a Lagrangian as it does not satisfy the Euler-Lagrange equation ($\star$). Funnily enough, it would've been simpler to not bother with $C$ at all, as $C$ is simply useless and only complicates the mathematics unless you manage to reduce the RHS to zero.
