Try the following article "A Survey of the Complemented Subspace Problem":
https://arxiv.org/abs/math/0501048v1
Your suspicion about $c_0$ is correct. A couple of other examples: The disc algebra (those functions in $C(\mathbb{T})$ which are restrictions of functions analytic in the open unit disc) is closed in $C(\mathbb{T})$ but not complemented. Similarly, in $L^1(\mathbb{T})$, the subspace $H^1(\mathbb{T})$ consisting of functions whose negative Fourier coefficients vanish is closed but not complemented. See Rudin's Functional Analysis (the proof isn't very easy).
Edit: I'm still not completely sure I understand your notation. For this answer, $F \oplus G$ means the algebraic sum of $F, G$, i.e $F \oplus G = \{x + y : x \in F, y \in G\}$, with the requirement that $F \cap G = \{0\}$, and $F = F_1 \oplus F_2 \oplus \dots$ means
$$F = \bigcup_{n \ge 1} F_1 \oplus \dots \oplus F_n.$$
The answer to your question is: not necessarily. For instance, it may be that $F_2 \oplus F_3 \oplus \dots$ is dense in $X$.
To be explicit, take $X = C([0,1])$. I don't quite understand the point of wanting the $F_n$ to have dimension larger than 1, and it will make this example a little messier, but for $n \ge 2$ let $F_n$ be the span of $x^{2n-4}$ and $x^{2n-3}$. Then $F_2 \oplus F_3 \oplus \dots$ is precisely the space $P$ of polynomials, which by the Weierstrass approximation theorem is dense in $X$. Let $g_1, g_2$ be bump functions supported in $[0,1/4], [3/4,1]$ respectively, and let $F_1$ be the span of $g_1$ and $g_2$. Every function in $F_1$ vanishes identically on $[1/4, 3/4]$ so if it is a polynomial it is 0. Thus $F_1 \cap P = \{0\}$, so $F_1 \oplus F_2 \oplus \dots$ is still a direct sum, and it is still dense in $X$ (since it contains $P$), so your hypotheses are satisfied. But the only $W$ that contains $\overline{F_2 \oplus F_3 \oplus \dots} = X$ is $X$ itself, and so $W \cap F_1 = F_1$.
Best Answer
No. For example $c_0$ is not complemented in its double dual $l^\infty$ (with detailed reasons here). But it is true that $X^*$ is always complemented in $X^{***}$. (and by this $c_0$ cannot be isomorphic to a dual space)