Easy cubes

Decomposing the factors in primes:

$3!=1\cdot 2\cdot 3$

$5!=1\cdot 2\cdot 3\cdot \left( 2\cdot 2 \right) \cdot 5$

$7!=1\cdot 2\cdot 3\cdot \left( 2\cdot 2 \right) \cdot 5\cdot \left( 2\cdot 3 \right) \cdot 7$

Multiplying the factorials and grouping the terms to create perfect cubes:

${ 1 }^{ 3 }\cdot { 2 }^{ 3 }\cdot { 3 }^{ 3 }\cdot { 2 }^{ 3 }\cdot { 2 }^{ 2 }\cdot { 5 }^{ 2 }\cdot { 3 }^{ 1 }\cdot { 7 }^{ 1 }$

Combining the factors to create other perfect cubes we have:

${ 2 }^{ 3 }\cdot { 3 }^{ 3 }={ 6 }^{ 2 }$

${ 2 }^{ 3 }\cdot { 2 }^{ 3 }={ 4 }^{ 3 }$

${ 2 }^{ 3 }\cdot { 2 }^{ 3 }\cdot { 3 }^{ 3 }={ 12 }^{ 3 }$

${ 2 }^{ 3 }\cdot { 1 }^{ 3 }={ 2 }^{ 3 }$

${ 3 }^{ 3 }\cdot { 1 }^{ 3 }={ 2 }^{ 3 }$

$1\cdot { 1 }^{ 3 }={ 1 }^{ 3 }$

In total we have $\boxed{6}$ perfect cubes that divides $3!\cdot 5!\cdot 7!$