If $n \geq 2$, then in fact you have $T = 0$. This follows from Fubini's theorem. Consider the two dimensional case. Denote by $\partial_c$ the classical derivatives operator and by $\partial_d$ the distributional derivatives, and let $\phi \in C^{\infty}_c(\Omega)$. Because $\frac{\partial_c f}{\partial x} \phi \in L^1(B)$ we have
$ \int_B \frac{\partial_c f}{\partial x} \phi dx dy = \int_{-1}^1 \int_{-1}^{1} \frac{\partial_c f}{\partial x}(x,y)\phi(x,y) dx dy = \int_{-1}^1 F(y) dy $ where $ F(y) = \int_{-1}^1 \frac{\partial_c f}{\partial x}(x,y)\phi(x,y) dx. $ Since for all $y \neq 0$, the function $x \mapsto \frac{\partial_c f}{\partial x}(x,y)$ is $C^1$, you can integrate by parts, to get $ F(y) = -\int_{-1}^1 f \frac{\partial_c \phi}{\partial x} dx. $ Applying Fubini again, we get $ \int_B \frac{\partial_c f}{\partial x} \phi dx dy = - \int_B f \frac{\partial_c \phi}{\partial x} dx dy$ which implies that $\frac{\partial_c f}{\partial x} = \frac{\partial_d f}{\partial x}$.
If $n = 1$, then the fundamental theorem of calculus together with the fact that $f \in L^1(B)$ implies that $f$ must have at most a jump discontinuity at $0$ and so you can indeed get only a constant multiple of a Dirac delta.
More generally, the Fubini argument applies to any $C^1$ function outside a singular set of "codimension" two or higher. For example, if $n = 3$ and $f \in C^1(B \setminus \{x = y = 0\})$, you can still run the argument. This is also why it fails for $n = 1$.