The relation $\dim(\mathrm{coker}(T))<\infty\iff\dim(\ker(T^*))<\infty$ remains true in arbitrary Banach spaces:
If $T$ has finite dimensional co-kernel then $\overline{\mathrm{im}(T)}$ admits a finite dimensional complement, choose one such complement and call it $V$. Note that $T^*(f)=0$ iff $f(Tv)=0$ for all $v\in X$, ie iff $f\lvert_{\overline{\mathrm{im}(T)}}=0$. As such $f$ is uniquely determined by its values on $V$, ie the map $\ker(T^*)\to V^*$ given by $f\mapsto f\lvert_V$ is injective. But $V^*$ is finite dimensional, so $\ker(T^*)$ also is finite dimensional.
On the other hand if the co-kernel is not finite dimensional then $\overline{\mathrm{im}(T)}$ admits infinite dimensional (not necessarily closed though) complements. Now you can check for any $V$ finite dimensional and linearly independent to $\overline{\mathrm{im}(T)}$ that for any $f\in V^*$ the map $V\oplus \mathrm{im}(T)\to\Bbb C, (v,x)\mapsto f(v)$ is continuous. In particular it admits Hahn Banach extensions with domain all of $X$. But any such Hahn Banach extension is $0$ on $\overline{\mathrm{im}(T)}$ hence lies in $\ker(T^*)$. Since you can do this for all $f\in V^*$ where $V$ has arbitrary finite dimension you must find that $\ker(T^*)$ is infinite dimensional.
In the case that $X$ is reflexive, I believe this is true, but I haven't checked it out; but I'm pretty sure that the map $B(X^*)\to B(X^{**})$ composed with the canonical isometric isomorphism $B(X^{**})\cong B(X)$ will give you the desired surjectivity.
However, the well-defined linear isometry $B(X)\to B(X^*)$, $T\mapsto T^*$ is not surjective in general. Take a non-reflexive Banach space $X$ (like $c_0$, or $\ell^1$) and let $j:X\to X^{**}$ be the canonical inclusion. Since $j$ is not surjective, let $\chi\in X^{**}$ be an element outside of $j(X)$. Fix a non-zero functional $\psi_0\in X^*$ and define an operator $P:X^*\to X^*$ by $P(\phi):=\chi(\phi)\cdot\psi_0$. This is a well-defined, bounded operator (and actually its range is one dimensional, but we dont care about it).
Now assume that $P$ is in the range of $\alpha:B(X)\to B(X^*)$, so there exists a bounded operator $T\in B(X)$ such that $P=T^*$, i.e. $P(\phi)=T^*\phi=\phi\circ T$ for all $\phi\in X^*$, so $\phi(Tx)=\chi(\phi)\cdot\psi_0(x)$ for all $x\in X$ and all $\phi\in X^*$. Since $\psi_0$ is non-zero, find $x_0\in X$ such that $\psi_0(x_0)\ne0$. We then have
$$\phi(Tx_0)=\chi(\phi)\cdot\psi_0(x_0)$$
and this is true for all $\phi\in X^*$; set $\psi_0(x_0):=\lambda\in\mathbb{C}\setminus\{0\}$. We have just shown that $\chi=\frac{1}{\lambda}j_{Tx_0}=j_{\frac{1}{\lambda}Tx_0}\in j(X)$, which is a contradiction.
Note that this answer works for any non-reflexive space; in other words, if the claim in my first paragraph is true (which i think it is) we get the following:
Corollary: Let $X$ be a Banach space. The canonical linear isometry $B(X)\to B(X^*)$, $T\mapsto T^*$ is surjective if and only if $X$ is reflexive.
Comment: I've spent a few hours thinking about this, so I have to give some credit to GEdgar. I knew the counter-example would come from the non-reflexive world from the first moment I started on this, but I was trying to give an answer specifically for $X=c_0$; after seeing GEdgar's comment I realized there is no need to go in a specific space, the abstraction actually helps here.
Best Answer
If $f\in Y^*$ then $f$ is a map from $Y$ to $\Bbb R$ or to $\Bbb C$. We define the map $g=T^*(f)$ from $X$ to $\Bbb R$ or to $\Bbb C$ by letting $g(x)=f(T(x))$ for each $x\in X.$ That is, $T^*(f)(x)=f(T(x)).$
Note that this is not $T^*(f(x)),$ which is meaningless. $T^*(f)$ is to be treated as a single symbol for one member of $X^*$.
For example if $X=Y=\Bbb R^2$ and $T(u,v)=(2v,u)$ for all $(u,v)$ then for any $f\in Y^*$ we have $T^*(f)(u,v)=f(T(u,v))=f(2v,u).$