Evidence that UniqueSat is dense
As far as I can tell, UniqueSAT is exponentially dense, in the sense that it contains $2^{\Omega(n)}$ instances of size $n$. (This is a stronger requirement than $2^{n^\varepsilon}$ for infinitely many $n$.)
The following argument works even for the revised definition of "dense" introduced in version 5 of the question, where all instances of size up to $n$ are taken into account.  Note that $\sum_{i=0}^n 2^i = 2^{n+1} - 1$.

Consider formulas in CNF with $v$ clauses, each clause containing exactly one literal, such that each variable occurs precisely once in the formula.
An example of such a "complete 1-SAT" formula for $v=3$ is $x_1\land x_2 \land \lnot x_3$, completely specifying the solution $x_1 = 1, x_2 = 1, x_3 = 0$.

We establish an exponential lower bound for the number of complete 1-SAT formulas of size $n$.
Logarithms are base 2.

Complete 1-SAT formulas each have precisely one solution, and with $v$ variables such a formula has size $v(1+\log v)$.
There are also $2^v.v!$ distinct complete 1-SAT formulas over $v$ variables.

We then just need to ensure that $v$ is as large as possible in terms of the instance size $n$.

Let $v = n/(\log n)^\delta$ for some $\delta = \delta(n)$.
For the size of complete 1-SAT formulas with $v$ variables to be precisely $n$ bits, we require that
$n = v(1+\log v)
= (n/(\log n)^\delta)(1+\log n - \delta\log\log n)$
so $1+\log n - \delta\log\log n = (\log n)^\delta$.
We can always choose $\delta$ appropriately to satisfy this expression; note that $1/2 &lt; \delta &lt; 1$ for large enough $n$, with $\delta \to 1$ as $n \to \infty$.

This already shows that there are at least $2^{n/\log n}$ such instances, so UniqueSAT is not sparse.

Moreover, the high order term of the exponent of $2^v.v!$ is then $n[(u/(1+u))-(\log e - 1)]$, where $u = \log n - \delta \log\log n$ and $\log e - 1 \approx 0.4427$.
(For instance, this follows from Robbins' bounds for the factorial function, $\log n! = n\log n - n\log e + 0.5\log n +r_n\log e + 0.5(1+\log \pi)$, where $1/(12n+1) &lt; r_n &lt; 1/(12n)$ for $n \ge 1$.)
Now $u/(1+u) &gt; 1/2$ for large enough $n$, so there are $2^{\Omega(n)}$ complete 1-SAT instances, and hence $2^{\Omega(n)}$ UniqueSAT instances.