Bjorn Poonen addresses this question for commutative (associative, unital) algebras in *<a href="https://arxiv.org/abs/math/0608491">The moduli space of commutative algebras of finite rank</a>*; 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 *<a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=1295&option_lang=eng">An estimate of the number of parameters defining an $e$-dimensional algebra</a>* 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. 

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**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 <a href="https://en.wikipedia.org/wiki/Heisenberg_group#Heisenberg_algebra">Heisenberg algebra</a> 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...