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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 $C \ne aT^b+k$)

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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.