If you're assuming $M$ is simply-connected, then it would be spin (since the Leech lattice is even). So a smooth manifold would violate Rokhlin's theorem. In the topological case (still assuming simple connectivity), $M$ would be oriented cobordant to a connected sum of three $E_8$ manifolds since they have the same signature and Kirby-Siebenmann invariant. (I think this is in the Freedman-Quinn book but I don't have it handy to check. An alternate reference is a paper of Hsu on the topological cobordism group.) So you could obtain it by a sequence of surgeries on that sum. This last fact uses the existence of handlebody structures on topological $5$-manifold pairs, which was proved by Quinn [J. Diff Geom. 17 (1982) 503-521] as a consequence of the annulus conjecture.
If you are not asking about the simply connected case, then the situation is a little more complicated. First, as Moishe noted in the comments, Donaldson's theorem would still rule out a smooth manifold. But in this setting the other question (about obtaining it by surgeries) might have a negative answer. I suspect (but don't know) that there is a non-spin manifold with form the Leech lattice, having fundamental group $\mathbb{Z}/2$, and trivial Kirby-Siebenmann invariant. (This would be analogous to the Enriques surface, which is smooth (so KS=0), non-spin, with fundamental group $\mathbb{Z}/2$, and form $E_8 \oplus$ hyperbolic.) I believe that such a manifold would not be cobordant to a sum of $E_8$ manifolds, which has non-zero KS invariant.
Added 05/30/23: The question asked about an explicit sequence of surgeries going from one manifold to another. I don’t think you can be completely explicit but you can say something. In the notation below I'll use the same letter for a manifold and its intersection form. First, find an explicit isomorphism $F$ between the forms $L \oplus H_1$ and $3 E_8 \oplus H_2$. Here $H_1,\ H_2$ are copies of a hyperbolic form of rank 2, generated by two classes, say $A_i$ and $B_i$.
Note that surgery on $L \# S^2 \times S^2$ along a 2-sphere representing $A_1$ produces a manifold with form $L$, and similarly for $3E_8 \# S^2 \times S^2$ (surgery $A_2$). So you get a sequence of surgeries
$$
L \overset{(1)}{\rightarrow} L \# S^2 \times S^2\overset{(2)}{\rightarrow} 3E_8
$$
as follows. (1) means do surgery on a circle (keeping things spin). This creates the class $A_1$, but surgery on that just gets you back to $L$. So (2) means do surgery on a $2$-sphere (with simply connected complement) representing $F^{-1}(A_2)$.
This is not really explicit, since you need Freedman's techniques to represent $F^{-1}(A_2)$ by a sphere. You'd also need to find $F$; this seems like work to me. The existence of such an isomorphism is a standard fact (classification of indefinite unimodular forms over $\mathbb{Z}$). But I don't know how to write down $F$.