I don't think that map is always surjective. For example, suppose that $Y = X$ is a supersingular elliptic curve. Then the Frobenius map $H^1(X, O_X) \to H^1(Y, O_Y)$ is the zero map (and both are 1-dimensional vector spaces). See for example Hartshorne's chapter on elliptic curves.
Alternately, suppose that $X$ is a scheme over $\mathbb{R}$, and $Y = X \times_{\mathbb{R}} \mathbb{C}$ is the base change. I don't think one should expect that $H^n(X, O_X) \to H^n(Y, O_Y)$ is basically ever surjective unless they are both zero...
With regards to your question though, here's one answer:
Suppose that $X$ is a normal integral scheme of characteristic zero and $Y$ is also integral. Then the natural map $O_X \to f_* O_Y$ splits as a map of $O_X$-modules, say with splitting map $\phi : f_* O_Y \to O_X$ (use the trace map on the fields $K(Y) \to K(X)$ and restrict to the structure sheaves). Now apply the functor $H^n(X, \bullet)$ to the composition (which is an isomorphism):
$$
O_X \to f_* O_Y \xrightarrow{\phi} O_X.
$$
Clearly one gets that
$$
H^n(X, O_X) \to H^n(X, f_* O_Y) = H^n(Y, O_Y) \to H^n(X, O_X)
$$
is also an isomorphism and thus
$$
H^n(X, O_X) \hookrightarrow H^n(Y, O_Y)
$$
injects as desired.
EDIT
Since the author of the question is particularly interested in the case when $f : Y \to X$ is the normalization of an $n$-dimensional $X$, let me try to say a couple things about that case.
Since we have a short exact sequence $0 \to O_X \to f_* O_Y \to C \to 0$, and since $f$ is birational, the support of $C$ has dimension $< \dim X$. Therefore, $H^n(X, O_X) \to H^n(Y, O_Y)$ is surjective as the original question states.
To show injectivity, it is sufficient to show that $H^{n-1}(X, C) = 0$. This will happen certainly if the non-normal locus of $X$ has codimension $\geq 2$. Otherwise, it generally won't happen.
A simple example with curves
Suppose that $X$ is a curve with exactly a node and $Y$ is its normalization (although any singular curve will work). Then $C$ is the skyscraper sheaf supported at a point. In particular $\dim H^0(X, C) = 1$. On the other hand, $\dim H^0(Y, O_Y) = 1 = \dim H^0(X, O_X)$, and so the exact sequence
$$
0 \to H^0(X, O_X) \to H^0(Y, O_Y) \to H^0(X, C) \to H^1(X, O_X) \to H^1(Y, O_Y) \to 0
$$
immediately implies that $H^0(X, C) = \ker H^1(X, O_X) \to H^1(Y, O_Y)$. In particular, the latter map is not injective.
Best Answer
There is a very general criterion for a map on $\pi_1$ to be surjective. Recall that for $X$ connected, the category of finite étale covers of $X$ is equivalent to the category $\pi_1(X)\text{ -}\operatorname{Set}_f$ of finite sets with a continuous $\pi_1(X)$-action. Under this correspondence, the $Y \to X$ finite étale with $Y$ connected correspond to the connected $\pi_1(X)$-sets $S$ (i.e. $\pi_1(X)$ acts transitively on $S$).
Lemma. Assume $X$, $Y$ connected, and $f \colon X \to Y$ a morphism. Then the induced morphism $\pi_1(f) \colon \pi_1(X) \to \pi_1(Y)$ is surjective if and only if for every $Z \to Y$ finite étale with $Z$ connected, the pullback $Z_X \to X$ is connected.
Proof. If $\pi_1(f)$ is surjective, then clearly any connected $\pi_1(Y)$-set is connected as $\pi_1(X)$-set. Conversely, if $\pi_1(f)$ is not surjective, then some $\gamma \in \pi_1(Y)$ is not in the image. Since fundamental groups are profinite, the image of $\pi_1(f)$ is closed, so the image of $\pi_1(f)$ misses some open neighbourhood of $\gamma$. Thus, there exists an open subgroup $U \subseteq \pi_1(Y)$ such that $$\gamma U \cap \operatorname{im} \pi_1(f) = \varnothing.$$ Then the finite $\pi_1(Y)$-set $S = \pi_1(Y)/U$ is not connected as $\pi_1(X)$-set. But it is clearly connected as $\pi_1(Y)$-set. $\square$
To apply this to the specific geometric setting you are interested in, just note that if $f \colon X \to Y$ has connected geometric fibres, then the same holds for the base change to any finite étale covering $Z \to Y$. It is then clear that if $Z$ is connected, so is $Z \times_Y X$.
Remark. There are more equivalent criteria for surjectivity; see for example Tag 0B6N. The one I gave above is amongst the ones listed, but this was not the case at the time of writing; hence my writing out the proof. My proof above was originally part of the proof of Tag 0BTX.