Bjorn Poonen addresses this question for commutative (associative, unital) algebras in The moduli space of commutative algebras of finite rank; asymptotically we have
$$q^{\frac{2}{27} n^3 + O(n^{8/3})}$$
such algebras (Theorem 10.9). Bjorn also gives a more precise lower bound on the dimension of the corresponding affine scheme in Theorem 9.2 which is a collection of three polynomials with leading term $\frac{2}{27} n^3$ depending on the value of $n \bmod 3$. The $\frac{2}{27}$ may seem familiar from a corresponding count of the number of finite $p$-groups and it happens for very similar reasons as he discusses in Section 10:
The approach towards both those results is to adapt the proof (begun in [Hig60] and completed in [Sim65]) that the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O(n^{8/3})}$. As suggested to us by Hendrik Lenstra, there is an analogy between the powers of the maximal ideal of a local finite-rank $k$-algebra and the descending $p$-central series of a $p$-group. Although there seems to be no direct connection between finite-rank $k$-algebras and finite $p$-groups, the combinatorial structure in the two enumeration proofs are nearly identical.
He also cites An estimate of the number of parameters defining an $e$-dimensional algebra by Yuri Neretin (which is in Russian, sadly for me) as addressing the Lie and associative cases; I'm not sure if the estimates immediately carry through to a finite field but if they do the answer is the same for Lie algebras and for associative algebras it is
$$q^{ \frac{4}{27} n^3 + O(n^{8/3}) }.$$
Presumably the analogous structure for Lie algebras making the answer similar is the descending central series for a nilpotent Lie algebra. For the associative case maybe it's something like powers of the Jacobson radical?
Note also that because $\frac{8}{3} > 2$ the error term in the exponent absorbs multiplicative factors as large as $q^{O(n^2)}$ so these asymptotics hold regardless of whether or not we quotient by the action of $GL_n(\mathbb{F}_q)$ (which is equivalent to asking for the isomorphism classification), which you might see as unsatisfactorily lenient but I think these are state of the art.
Edit: The lower bound for Lie algebras is easy enough to give here; it's very similar to the argument for finite $p$-groups and for commutative algebras but, I think, simpler. We consider only 2-step nilpotent Lie algebras $L$ of some dimension $n$, which arise as a central extension
$$0 \to [L, L] \to L \to A \to 0$$
of an abelian Lie algebra $A$ (the abelianization) by another abelian Lie algebra $[L, L]$ (the commutator; I am not using fraktur here to save typing). Explicitly, the Lie bracket $[-, -]$ factors through $A$ and lands in $[L, L]$, and so the only constraint on it is that it's a surjective alternating map $\wedge^2(A) \to [L, L]$; given any such map we can construct a Lie bracket which trivially satisfies the Jacobi identity because all triple commutators vanish by 2-step nilpotence. This is a mild generalization of the construction of the Heisenberg algebra where $\dim [L, L] = 1$.
So, fixing the vector space $L$, we put a 2-step nilpotent Lie algebra structure on $L$ by first choosing a subspace $[L, L]$ we want to be the commutator and then choosing a surjection $\wedge^2(L/[L, L]) \to [L, L]$. In general the space of surjections from a f.d. vector space $V$ to a f.d. vector space $W$ admits a free action by $GL(W)$ and the quotient by this action is the Grassmannian of codimension $\dim W$ subspaces of $V$. So, setting $b = \dim [L, L]$, the space of choices we have available is the triple of choices of
- a $b$-dimensional subspace $[L, L]$ of $L$,
- a $b$-codimensional subspace of $\wedge^2(L/[L, L])$, and
- an isomorphism between the first choice and the quotient by the second choice.
Write $a = n - b = \dim L/[L, L] = \dim A$, so that $a + b = n$. Over $\mathbb{F}_q$ there are exactly
$${n \choose b}_q { {a \choose 2} \choose b}_q |GL_b(\mathbb{F}_q)|$$
ways to make the above choices. Now our job is to find $a, b$ which maximizes this, or at least which makes it quite big since we're aiming for a lower bound. The leading term in $q$ (which is a lower bound since all the coefficients are positive!) is $q$ to the power of
$$ab + \left( {a \choose 2} - b \right) b + b^2 = \frac{a(a+1)b}{2}.$$
Subject to the constraint that $a + b = n$ this is maximized when $a \approx \frac{2n}{3}, b \approx \frac{n}{3}$, and we could be more careful depending on the value of $n \bmod 3$ if desired. Let me instead restrict to the case that $3 \mid n$ so that we can divide by $3$ exactly, so our leading term and lower bound is that there are at least
$$q^{ \frac{2}{27} n^3 + \frac{2n^2}{3} }$$
2-step nilpotent Lie algebra structures on a finite-dimensional vector space $L$ of dimension $3 \mid n$ over $\mathbb{F}_q$. You didn't ask for this, but I want to mention it anyway: quotienting by the action of $GL_n(\mathbb{F}_q)$ to get isomorphism classes at worst divides this by $q^{n^2}$, so we get that there are at least
$$q^{ \frac{2}{27} n^3 - \frac{n^2}{3} }$$
isomorphism classes of 2-step nilpotent Lie algebras of dimension $3 \mid n$ over $\mathbb{F}_q$, with similar explicit bounds for $n \equiv 1, 2 \bmod 3$. It is maybe surprising that it's possible to prove a matching upper bound, at least up to leading order in the exponent; I don't know what that argument looks like in detail. Plugging in some actual numbers, we get that there are
- at least $q^8$ Lie brackets on $\mathbb{F}_q^3$ and so at least $q^{-1}$ isomorphism classes of $3$-dimensional lie algebras (technically true!)
- at least $q^{40}$ Lie brackets on $\mathbb{F}_q^6$ so at least $q^4$ isomorphism classes of $6$-dimensional Lie algebras, and
- at least $q^{108}$ Lie brackets on $\mathbb{F}_q^9$ so at least $q^{27}$ isomorphism classes of $9$-dimensional Lie algebras.
For small values of $n$ it would be feasible to not only maximize but sum over all $a + b = n$ above and so compute the exact number of 2-step nilpotent Lie brackets. The truly brave who wanted to compute the exact number of isomorphism classes could hope to apply Burnside's lemma...
The lower bound for finite groups is very similar, using 2-step nilpotent $p$-groups arising as the central extension of a f.d. $\mathbb{F}_p$-vector space by another one, etc.