You basically have the idea - note that the quadratic residues $\mod{5},\mod{8}$ are both $\{0,1,4\}$, and you need to find $n$ such that $(2n+1),(3n+1)$ are both amongst these for each modulus.
Checking cases gives that $(2n+1)\mod{5}$ and $(3n+1)\mod{5}$ are both in $\{0,1,4\}$ only when $n=0\mod{5}$, and the same calculations $\mod{8}$ tell you that $n$ must be $0\mod{8}$.
Thus, $n$ is a multiple of $5$ and $8$, which are coprime, hence $n$ is a multiple of $5\times8=40$.
Partial solution:
Let $a=111\dots112=b^3$. Then $b^3-1=(10^c-1)/9$ for some integer $c$. Rewrite this as $10^c=9b^3-8$. We split into two cases, depending on the parity of $c$.
Case 1. If $c$ is even, then $10^c$ is a square, and we have $d^2=9b^3-8$ (where $d$ has to be a power of $10$). Multiplying through by $81$, we get
$$
y^2=x^3-648,
$$
where $y=9d$ and $x=9b$. This is the Mordell equation $y^2=x^3-k$, with $k=648$. Solutions of Mordell equations are tabulated at https://hr.userweb.mwn.de/numb/mordell.html#tbl5 where we find there are six solutions,
$$
(x,y)=(9,9),(18,72),(22,100),(54,396),(97,955),(1809,76941)
$$
Only the first of these satisfies the condition that $y$ be nine times a power of ten, and $(x,y)=(9,9)$ doesn't lead to a cube of the requested form, so we get no solutions when $c$ is even.
Case 2. If $c$ is odd, then $10^c$ is ten times a square, and $10d^2=9b^3-8$. By manipulations similar to those in Case 1, this leads to the Mordell equation
$$
y^2=x^3-648000
$$
Unfortunately, the tables don't go that high, so we'd actually have to put in some work to find all the solutions, if any, to this equation. Perhaps someone else will take up the problem from here.
Best Answer
There are no solutions. As noted by others, the amounts to solving the equation $$ 55 \cdot 10^n-9y^3=1. $$ Writing $n=3n_0+k$, for $k \in \{ 0, 1, 2 \}$, this leads, crudely, to $3$ Thue equations of the shape $$ Ax^3-9y^3=1, \; \; A \in \{ 55, 550, 44 \}. $$ In each case, we can use, for example, Pari GP or Magma to solve the equation and show that there are no solutions (for cubic equations of this shape, it's somewhat easier than usual due to old work of Nagell which shows that any possible solutions correspond in a very precise sense to fundamental units in the associated cubic fields).