Let $f:=\mathbf{a}^{T}\mathbf{x}$. The claim that: $$\nabla f=\nabla\mathbf{a}^{T}\mathbf{x}=\|\mathbf{a}\|_{1}=a_{1}+a_{2}+\cdots+a_{n}\tag{1}$$ is false where in fact the true answer is $\mathbf{a}^{T}$.
Now one would argue that the correct answer is true by simply considering $\mathbf{a}$ to be just a scalar). However, I ended up proving $(1)$ and therefore I would request someone to indicate for me where in my attempt I made a mistake: \begin{align} \nabla f&=\frac{\mathrm{d}}{\mathrm{d}\mathbf{x}}\mathbf{a}^{T}\mathbf{x}\tag{2}\\ &=\frac{\mathrm{d}}{\mathrm{d}\mathbf{x}}\sum_{i=1}^{n}a_{i}x_{i}\tag{3}\\ &=\sum_{i=1}^{n}a_{i}\frac{\mathrm{d}}{\mathrm{d}x_{i}}x_{i}\tag{4}\\ &=\sum_{i=1}^{n}a_{i}\tag{5}\\ &=\|\mathbf{a}\|_{1}\tag{6} \end{align}
Using Einstein summation, $$\boldsymbol{a^\top x}=a_ix^i \\ \nabla_j(a_ix^i)=a^i\nabla_j x^i \\ =a^i\delta^i_j \\ =a_j$$ $a_j$ is the $j$th entry in the index-lowered or transposed version of $\boldsymbol a$. So $$\nabla(\boldsymbol{a^\top x})=\boldsymbol {a^\top}$$
Your mistake was writing $$\nabla \overset{?}{=}\frac{\mathrm d}{\mathrm d\boldsymbol x}\neq \sum_i \frac{\mathrm d}{\mathrm dx^i}$$ It is in fact $$\nabla=\sum_i\frac{\partial}{\partial x^i}\boldsymbol e_i$$