I know that there are manifolds that are not separable, such as the long line $\omega_1 \times [0, 1) \setminus \{(0, 0)\}$. Is there any manifold with no uncountable dense set of cardinality $\aleph_1$?
A line $\omega_2 \times [0, 1) \setminus \{(0, 0)\}$ wouldn't work since no neighborhoods of $(\omega_1, 0)$ are homeomorphic to $\mathbb{R}$. Would something else work?
On
For dimension $1$, I believe the answer is no.
Suppose $M$ is a connected Hausdorff space locally homeomorphic to $\mathbb{R}$. Then we get a nondegenerate betweenness relation on $M$: $x$ is between $a$ and $b$ if every connected open set containing $a$ and $b$ also contains $x$. This in turn lets us define intervals: $[a,b]$ is the set of all $x$ which are between $a$ and $b$.
Now let $\langle x\rangle$ be the union of all intervals $[a,b]$ such that $x$ is between $a$ and $b$ and $[a,b]$ is homeomorphic to the usual unit interval. It's easy to check that $\langle x\rangle$ and $M\setminus\langle x\rangle$ are each open; since $M$ is connected this means $\langle x\rangle=M$.
But now consider a maximal sequence $\mathfrak{I}=(I_\eta)_{\eta<\alpha}$ of intervals containing $x$ such that for each $\eta_1<\eta_2<\alpha$ we have $I_{\eta_1}\subsetneq int(I_{\eta_2})$. Such a sequence can have length at most $\omega_1$ since otherwise we get an order-preserving injection of $\omega_1$ into the unit interval. And this gives us a dense subset of $\langle x\rangle=M$ of size $\le\omega_1$.
It is at least consistent that every manifold have a dense subset of cardinality $\omega_1$: every manifold has cardinality $2^\omega=\mathfrak{c}$, so under $\text{CH}$ every manifold has cardinality $\omega_1$. (This is Theorem 2.9 in Peter Nyikos, The Theory of Nonmetrizable Manifolds, in Handbook of Set-Theoretic Topology, K. Kunen & J.E. Vaughan, eds., North Holland, 1984.) However, in section 3.7 he constructs a manifold of cellularity (and therefore density) $\mathfrak{c}$; under $\neg\text{CH}$ it has no dense subset of cardinality $\omega_1$.