As Qiaochu notes, kernels and cokernels don't make sense in an arbitrary category: you need a zero object to even define what these should be (and even with a zero object, they aren't guaranteed to exist).
If you have a category with a zero object, then a monomorphism $f$ has zero kernel: $f\circ\ker f = 0 = f\circ 0\implies \ker f = 0$. This doesn't show that the kernel actually exists, just that if it does, it must be equal to zero. However, it is not difficult to verify that $0$ satisfies the universal property of the kernel in this case. Reverse all the arrows to get the dual statement for cokernels.
However, the converse is false: consider the category with four objects $A$, $B$, $C$, and a zero object $0$ and morphisms
\begin{align*}
\operatorname{Hom}(X,X) &= \{0, id_X\}\quad\textrm{for each object }X\\
\operatorname{Hom}(B,C) &= \{0, f\}\\
\operatorname{Hom}(A,B) &= \{0,g,h\}\\
\operatorname{Hom}(A,C) &= \{0, fg = fh\}\\
\operatorname{Hom}(C,A) &= \operatorname{Hom}(C,B) = \operatorname{Hom}(B,A) =\{0\}.
\end{align*}
By construction, $g\neq h$, $fg = fh$, and you can verify that $\ker f$ exists and is equal to $0 : 0\to X$.
Best Answer
The other answer deals with monomorphisms, so let's look at epimorphisms. As pointed out in comments, there is a (1970?) preprint of G. Bergman's, Epimorphisms of Lie Algebras, available online, which addresses this question. Further, in the notes at the end of that preprint, we find a reference to Reid, G.A.: Epimorphisms and Surjectivity. Inventiones mathematicae, Volume 9 (1969) pp. 295-307, also available online, which has an overlap with Bergman's preprint.
The highlights relevant to your question about epimorphisms are as follows. Note first of all that w.l.o.g. we can reduce to the question for what subalgebras $\mathfrak h \subseteq \mathfrak g$, the natural inclusion is an epimorphism.
If $K$ is any field, then in the category of all $K$-Lie algebras, necessarily $\mathfrak h =\mathfrak g$, i.e. epimorphisms are surjective (Reid Prop. 4; Bergman Thm 2.1). Either proof goes through the universal enveloping algebra and uses the Poincaré-Birkhoff-Witt theorem; Bergman does it from there with certain beautiful ring-theoretic characterisations of epimorphisms (for which cf. answers to MO/120918), whereas Reid uses the beautiful criterion that the inclusion $\mathfrak h \subset \mathfrak g$ is an epimorphism if and only if for every $\mathfrak g$-module $V$, every element $v \in V$ that is annihilated by $\mathfrak h$ is annihilated by all of $\mathfrak g$. And for $\mathfrak h \subsetneq \mathfrak g$, he constructs (via the universal enveloping algebra and PBW) a module $V$ where that is not the case.
If $char(K)=0$, then in the category of finite dimensional $K$-Lie algebras, there are epimorphisms which are not surjective. (Bergman Example 4.1, Reid Prop. 7). Indeed in analogy to the criterion above, Bergman shows (Corollary 3.2) that the inclusion $\mathfrak h \subseteq \mathfrak g$ is an epimorphism in this category if and only if for every finite dimensional $\mathfrak g$-module $V$, every element $v \in V$ that is annihilated by $\mathfrak h$ is annihilated by all of $\mathfrak g$; and then there are obvious examples of proper inclusions like this. Namely, as both sources note, if $\mathfrak g$ is split semisimple (like $\mathfrak{sl}_n(K)$, or every semisimple Lie algebra if $K=\mathbb C$), then the inclusion of any Borel subalgebra (and hence, of any parabolic subalgebra) $\mathfrak h \subset \mathfrak g$ is an epimorphism. Indeed, that Bergman's example, the inclusion of the standard Borel $$\{\pmatrix{\alpha & \beta \\0&-\alpha}:\alpha, \beta \in K\} \hookrightarrow \mathfrak{sl}_2(K) $$ satisfies the mentioned criterion for finite-dimensional $\mathfrak{sl}_2(K)$-representations, follows immediately from the basics of the representation theory of $\mathfrak{sl}_2$ covered in any worthwhile resource on Lie algebra representations.
Bergman points out various examples of non-surjective epimorphisms in the category of finite-dimensional Lie algebras over a characteristic $0$ field which are not immediately given by parabolic subalgebras as above, but notes that there still might be a chance to classify them, with the parabolic ones as a cornerstone (see p. 13/14 and "Addenda" at the end of Bergman's preprint). Correspondingly, Reid proves (Prop. 10) that in the category of real compact (fin.-dim.) Lie algebras, epimorphisms are surjective. (Note that real compact Lie algebras basically have no proper parabolic subalgebras.)
Bergman has further interesting results in positive characteristic, which for once seems to be better behaved than characteristic $0$. -- Reid, on the other hand, has the characteristic-free result (Prop. 5/6) that in the category of finite-dimensional nilpotent Lie algebras over $K$ as well as in the category of finite-dimensional solvable Lie algebras over $K$, epimorphisms are surjective.
So this was the state of the art 50 years ago. I would not be surprised, but delighted, to hear of more recent results especially regarding no. 3.