Over at MO, Bill Johnson have an answer which works for any Banach space with a Shrinking Basis. As no-one has given an answer here yet, I thought I'd show a more "simple minded" calculation which works in the special case of $c_0$.
Notice that the sum $ \sum_n T_{mn} a_n $ only converges, for all $(a_n)\in c_0$, if $\sum_n |T_{mn}|<\infty$. But this is not enough to ensure that $T$ is a bounded operator into $c_0$; for that, you need precisely that $ \sup_m \sum_n |T_{mn}|<\infty. $ (Edit: Actually, this only ensures that $T$ maps into $\ell^\infty$. For $T$ to map to $c_0$, we need that the rows of the matrix of $T$, treated as a sequence in $\ell^1$, is weak$^*$-null, equivalently, that each column of the matrix tends to $0$.) This quantity is $\|T\|$. So we have characterised which matrices can occur.
Now let $(f_m)\in\ell^1$, so $ \langle (f_m) , T(a_n) \rangle = \sum_m \sum_n f_m T_{mn} a_n. $ The double sum is absolutely convergent, so we can re-arrange, and conclude that $ T^*(f_m) = \Big[ \sum_m T_{mn} f_m \Big]. $ That is, $T^*$ is matrix multiplication by the tranpose of $T$ (as we might hope!)
So finally for $(x_n)\in\ell^\infty$, $ \langle T^{**}(x_n) , (f_m) \rangle = \sum_n x_n \sum_m T_{mn} f_m. $ Again, the double sum converges absolutely as $ \sum_{n,m} |T_{mn}| |f_m| |x_n| = \sum_m |f_m| \sum_n |T_{mn}| |x_n| \leq \|f\|_1 \sup_m \sum_n |T_{mn}| |x_n| $ $\leq \|f\|_1 \sup_m \|x\|_\infty \sum_n |T_{mn}| = \|f\|_1 \|x\|_\infty \|T\|. $ So we can re-arrange and conclude that, yes, $ T^{**}(x_n) = \Big[ \sum_m T_{mn} x_n \big]. $ In particular, for fixed $m$, the sequence $(T_{mn})$ is in $\ell^1$, and so the sum genuinely converges.