Not a full answer but too long for a comment.
We rewrite the problem as follows: Define $A:\mathbb{R}^n\to L^1(0,1)$ by $Aa(t) = \sum_k a_k\sin(2\pi kt)$. Then the problem is
$$
\min_a \|f-Aa\|_1
$$
with $f(t) = \sin(2\pi t)$
and this is a convex minimization problem in finite dimensions. A necessary and sufficient condition for $a^*$ to be optimal is
$$
0\in A^*\operatorname{Sign}(f-Aa^*)
$$
where $\operatorname{Sign}$ is the multivalued sign, i.e.
$$
\operatorname{Sign}(f(x)) = \begin{cases}\{1\} & f(x)>0\\ [-1,1] & f(x)=0\\\{-1\} & f(x)<0\end{cases} .
$$
The adjoint operator $A^*:L^\infty(0,1) \to \mathbb{R}^n$ is
$$
(A^*g)_k = \int_0^1 g(t)\sin(2\pi kt)dt.
$$
In other words: $a^*$ is optimal if there is a $g\in L^\infty(0,1)$ such that
i) $g(t) \in \operatorname{Sign}(f(t)-Aa^*(t))$ almost everwhere, and
ii) $\int_0^1 g(t) \sin(2\pi kt)dt = 0$ for all $k$.
I did not try to solve this for your actual $f$, though.
In fact, approximation in $L^1$ is a well studied subject and if you search for "trigonometric approximation in L^1" you'll find the paper "Interpolation and L1-approximation by trigonometric polynomials and blending functions" which may be helpful.