We know $x \neq 0$ and $\lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor \neq 0$, since their product is nonzero. Therefore we may divide both sides by the nonzero integer $n = \lfloor x \lfloor x \lfloor x \rfloor \rfloor \rfloor$, and we get that $x = \frac{88}{n}$ is rational. So $x = \frac{p}{q}$ for some integers $p, q$, where $q \neq 0$.
Further, $\frac{88}{n} = \frac{p}{q}$ where $\frac{p}{q}$ is in lowest terms, so $p$ is a divisor of $88$. Since $x > 0$, we may assume $p>0$ and $q >0$. Thus $p$ is one of $1, 2, 4, 8, 11, 22, 44, 88$. On the other hand, since $\lfloor x \rfloor > 0$, we have that $1\leq q \leq p$.
This narrows down the choices to $(p, q)$; there are now only $1 + 2 + 4 + 8 + 11 + 22 + 44 + 88 = 15 + 15(11) = 15(12) = 180$ possible pairs to check. (Actually fewer than that, since the same rational number may show up as $p/q$ for various different pairs $(p, q)$.)
The following Python 2.7 code returns the pairs $(p,q) = (22, 7)$, $(44, 14)$, and $(88,28)$, all of which represent fractions that reduce to $22/7$. Note that in Python 2.7, regular division of integers returns the integer quotient (aka the floor of the rational number).
for p in [1, 2, 4, 8, 11, 22, 44, 88]:
for q in xrange(1, p): #no need to check p = q since p/q = 1 is not a solution
frac = 1
for i in xrange(3):
frac *= p
frac /= q
if frac*p == 88*q:
print p,q
In fact, many of these same comments still apply even if we don't assume $x>0$. If we had assumed $x<0$ instead, then clearly $x \leq -1$: Otherwise, $-1 < x < 0$, so $\lfloor x \rfloor = -1$, and in that case $\lfloor x \lfloor x \rfloor \rfloor = \lfloor -x \rfloor = 0$ so the whole product on the left-hand side is $0$. Further, $x = -1$ is not a solution, so $x < -1$. Therefore $x = p/q$ where $p$ is a (positive) divisor of $88$ and $q$ is a (negative) integer such that $-p < q \leq -1$.
Checking these possible pairs yields no solutions, so $x = 22/7$ is the only real solution, positive or negative.