The expression you've written derives from a proposal, due to Ryu and Takayanagi (See: http://arxiv.org/abs/hep-th/0603001), to calculate entanglement entropy of a CFT by making use of a gravity dual.
Define a CFT in $D$ space-time dimensions. Take a spatial slice, a ball of radius $R$ for example, and calculate the entanglement entropy associated with it. You'll get the following behaviours:
$S_{D=2}(R)=\frac{c}{3}\log \frac{R}{\epsilon}$
$S_{D=3}(R)=k_1 \frac{R}{\epsilon} -F$
$S_{D=4}(R)=k_2 \frac{R^2}{\epsilon^2} -a \log \frac{R}{\epsilon}$
where $c$, $F$, $a$ are universal terms. For the case of $D=2$ it's the central charge.
Now if this CFT admits a holographic dual then there's another expression for the entanglement entropy:
$S(R) = \frac{A_{\gamma(R)}}{4G_N}$
where $G_N$ is Newton's constant and $A_{\gamma(R)}$ is the area of a minimal surface $\gamma(R)$ in $AdS_{D+1}$ whose boundary is exactly the boundary of the spatial region you're computing $S(R)$ for. Like this picture: 
The intuition for why this is true is that tracing out a region is like inducing a horizon beyond which you don't know what's going on. Then this holographic expression is like the Bekenstein-Hawking formula.
Matching expressions can give you a formula like you desire. An important thing to note though is that this $a$ for the case of $D=4$ isn't always the "$c$" central charge. (See my comment for a bit more on this) They differ in cases where your CFT is dual to a higher curvature gravity theory.(http://arxiv.org/abs/hep-th/9904179) This is related to its appearance in the trace anomaly.
This $a$ quantity does fulfill a role in generalizing the $c$-theorem of $D=2$ CFT to higher dimensions though; it's an RG monotone. This line of inquiry has been pursued by Rob Myers and Aninda Sinha. (See: http://arxiv.org/abs/1011.5819 and http://arxiv.org/abs/1006.1263)
EDIT:
As promised this will provide a formula for a 'central charge', or combination of them, in any number of dimensions which is just a function of constants and the AdS radius $R$.
For $D=4$ this yields: $a=\frac{\pi^{3/2}}{4 G^{(5)}_N} \frac{R^3}{\Gamma(\frac{3}{2})}$