Let $K$ be the ground field. An extension of a Lie $K$-algebra $\mathfrak{g}$ means a pair $(\mathfrak{h},p)$ where $p$ is a surjective homomorphism $\mathfrak{h}\to\mathfrak{g}$. A homomorphism between extensions $(\mathfrak{h}_1,p_1)$, $(\mathfrak{h}_2,p_2)$ is a $K$-algebra homomorphism $f:\mathfrak{h}_1\to\mathfrak{h}_2$ such that $p_2\circ f=p_1$. A universal extension of $\mathfrak{g}$, is naturally defined as an initial object in this category.
In the full subcategory of central extensions, an initial object is called a universal central extension (and unlike the claim in a comment to this answer, is a well-adapted definition). This is classical, as well as the fact that a universal central extension of $\mathfrak{g}$ exists if and only if $\mathfrak{g}$ is perfect.
In the whole category of extensions of $\mathfrak{g}$, one has:
Then there is no universal extension of $\mathfrak{g}$ except for $\mathfrak{g}=\{0\}$.
This holds in the category of Lie $K$-algebras, or also, if $\mathfrak{g}$ is finite-dimensional, in the category of finite-dimensional Lie $K$-algebras.
Since $\{0\}$ is an initial object in the category of Lie algebras, $\{0\}\to\{0\}$ is a universal extension. Hence suppose $\mathfrak{g}\neq\{0\}$. By contradiction, let $(\mathfrak{h},p)$ be a universal extension, so $\mathfrak{h}\neq\{0\}$. Consider the first projection $q:\mathfrak{g}\times\mathfrak{h}\to\mathfrak{g}$. Then both homomorphisms $(p,0)$ and $(p,\mathrm{id})$ from $\mathfrak{h}$ to $\mathfrak{g}\times\mathfrak{h}$ lift $p$, and this contradicts the uniqueness in the universal property.
(Note: this is a variation of the argument to show that if $\mathfrak{g}$ is not perfect then it has no universal central extension.)
Edit:
Let $\mathcal{C}$ be a full subcategory of the category of abelian extensions of $\mathfrak{g}$, not consisting of central extensions. Then there is a no universal $\mathcal{C}$-extension (i.e., no initial object in $\mathcal{C}$).
In particular, for $\mathfrak{g}\neq\{0\}$, there is no universal abelian extension of $\mathfrak{g}$.
Proof: let $(\mathfrak{h},p)$ be a universal extension in $\mathcal{C}$. Let $(\mathfrak{s},q)$ be non-central extension in $\mathcal{C}$, with kernel $Z$. Let $f:\mathfrak{h}\to\mathfrak{s}$ be given by the universal property. Choose $z\in Z$ not central; then $\tau:s\mapsto s+[s,z]$ is an automorphism of $\mathfrak{s}$, inducing the identity of $\mathfrak{s}$ modulo $s$. By the uniqueness in the universal property, it follows that $s\circ f=f$. Hence $f$ is valued in the set of points fixed by $s$, which is the centralizer of $z$. Since $q\circ f$ is surjective, we also have $f(\mathfrak{h})+Z=\mathfrak{s}$. Combining, we deduce that $z$ is central, contradiction.
(The second assertion follows since every nontrivial $\mathfrak{g}$ has a nontrivial $\mathfrak{g}$-module, hence has a non-central abelian extension.)
