See this question for Riemann--Hurwitz in higher dimensions.
But there cannot be a simple relation between (topological) Euler characteristic and "genus" in higher dimensions, for any definition of "genus" that is a function only of the Hodge numbers $h^{i,0}=h^0(X, \Omega^i)$. (Remember that for a curve, $g=h^{1,0}$.) The reason is that the Hodge numbers are birational invariants of smooth varieties. In particular, they don't change under blowing up. But the Euler characteristic does: for example, blowing up a point on a smooth surface increases the Euler characteristic by 1.
On the other hand, as Shafarevich goes on to say in the next paragraph, for any smooth projective variety $X$, there are some relationships between these Hodge numbers and the topology of $X$. For example the first Betti number of $X$ – that is, the rank of $H_2(X)$ – is $b_1(X)=2h^{1,0}$.
If you want a more precise relationship, you need to take into account all the Hodge numbers of $X$. (Recall that these are defined as $h^{i,j}=H^j(X,\Omega^i)$, so they involve not just holomorphic differentials, but higher cohomology of the sheaves of differentials. Don't ask me why $i$ and $j$ get flipped – read @GeorgesElencwajg's comment below!)
Then the Hodge Decomposition Theorem says that for each $k$, we have
$$ H^k(X,\mathbb C) \cong \bigoplus_{i+j=k} H^j(X, \Omega^i) \ ;$$
taking the alternating sum for $k$ from $0$ to $\operatorname{dim} X$ we get finally an expression for the Euler characteristic in terms of (cohomology groups of sheaves of) differentials on $X$:
$$e(X) = \sum_k (-1)^k b_k = \sum_{i,j} (-1)^{i+j} h^{i,j}.$$
To complete the picture, note that the only nonzero Hodge numbers on a curve are $h^{0,0}=h^{1,1}=1$ and $h^{0,1}=h^{1,0}=g$, so in this case the formula reduces to the familiar $e(X)=2-2g$ .
the sheaf $\Omega^i$ has a Dolbeault resolution $(...\to\mathcal E^{ij}\to...)j$ by fine sheaves which calculate its cohomology. So $H^j(X,\Omega ^i)$ is calculated by taking closed forms in $\mathcal E^{ij}(X)$ modulo exact forms . Since the forms in $\mathcal E^{ij}(X)$ are written as $\Sigma f{KL}dz^K \wedge d\bar z^L$ with $i$ factors $dz^k$ and $j$ factors $d\bar z^l$ this explains (I think) the flipped terminology $h^{ij}(X)$ . – Georges Elencwajg Jan 20 '15 at 14:46