Can the long line be embedded in euclidean space?

No, because of the Poincaré-Volterra theorem. Here's the statement you can find in Bourbaki's General Topology (I.11.7, corollary 3).

Theorem. Let $Y$ be a locally compact, locally connected space whose topology has a countable base. Let $X$ be a connected Hausdorff space, and let $p : X \to Y$ be a continuous mapping which has the following property: for each $x \in X$ there is an open neighbourhood $U$ of $x$ in $X$ such that the restriction of $p$ to $U$ is a homeomorphism of $U$ onto an open subspace of $Y$. Then $X$ is locally compact and locally connected, and the topology of $X$ has a countable base.

A classical use of this theorem is to show that Riemann surfaces are automatically secound countable, even if you don't make it an explicit hypothesis (the general theory shows that there is a nonconstant holomorphic map $S \to \mathbb P^1(\mathbb C)$ and you apply the Poincaré-Volterra theorem).

So, for a manifold, being secound-countable is equivalent to being embeddable in some Euclidean space. These are two of the 119 (!) equivalent conditions you can find on theorem 2.1 of Gauld's Non-metrisable manifolds.

Recall that every subspace of a second-countable space is also second-countable. (If $\mathcal B $ is countable base for $ X$ then $\{ U \cap Y : U \in \mathcal B \}$ is a countable base for $ Y \subseteq X$.) Therefore no non-second-countable space (manifold or otherwise) can embed in any Euclidean space.