As Sandor and Martin pointed out above, it is ok if the subschemes intersect in the emptyset pairwise.  I'm going to provide 3 examples, I think the third one gives a counter-example to even the transversality statements.

**Example 1:**  Here's an example where it's false without that hypotheses, notice that the varieties are smooth and they intersect *pairwise* with normal crossings.  **EDIT:** as t3suji pointed out in the comments, these varieties don't intersect transversally in the ambient space, just in the ambient $z = 0$ plane **EndOfEdit**  Consider $X = \mathbb{A}^3$ and set $Z_1 = V(y,z)$, $Z_2 = V(x,z)$, $Z_3 = V(x, z-1)$.  Notice that $Z_3$ doesn't intersect any of the other subschemes.

Then, $I_1 \cap I_2 = (z, xy)$.

However, $I_1 \cdot I_2 = (xy, yz, xz, z^2)$.

These ideals are not equal clearly.  Now, we can immediately see that multiplying/intersecting by $I_3$ won't change the behavior at the origin at all since the ideal doesn't vanish there, so they are not equal.  However, just to be sure, I also did the following computation (with Macaulay2):

$$I_1 \cdot I_2 \cdot I_3 = (xz, yz, z^3 - z^2, yz^2 - yz, xz^2 - xz).$$

$$I_1 \cap I_2 \cap I_3 = (z^2 -z, xz, xy).$$

Macaulay2 also confirmed that the ideals were not equal.

**Example 2:**  Here's a different example.  a 4th variety that doesn't intersect the others at all.
$X = \mathbb{A}^3$.  $I_1 = (x,y)$, $I_2 = (x,z)$, $I_3 = (y,z)$ and $I_4 = (x-1,y-1,z-1)$.  Certainly again the $I_4$ doesn't matter, it's just included so that the sum of the ideals is equal to $R = k[x,y,z]$.  I wonder if it might be reasonable to say that $Z_1$, $Z_2$ and $Z_3$, as a triple, have transverse intersection at the origin.  Anyways:

$I_1 \cap I_2 \cap I_3 = (xy, xz, yz)$ but,
$I_1 \cdot I_2 \cdot I_3 = (yz^2, xz^2, y^2z, xyz, x^2z, xy^2, x^2y)$.

**Example 3:**  Ok, now I'm just going to give three subvarieties which intersect at the origin, pairwise transversally, and such that the product of the ideals is not equal to the intersection.  You may add the ideal sheaf of some other variety that doesn't intersect them at all to make the sum of ideals equal the whole structure sheaf.

$X = \mathbb{A}^4 = \text{Spec}k[x,y,u,v]$.  $I_1 = (x,y)$, $I_2 = (u,v)$, $I_3 = (x+u, y+v)$.  I believe these have pairwise transverse intersection at the origin.  Now then, it is true that $I_1 \cdot I_2 = I_1 \cap I_2$, and likewise with any pair.  However, Macaulay2 can be used to verify that $I_1 \cdot I_2 \cdot I_3 \neq I_1 \cap I_2 \cap I_3$.  Roughly speaking the problem is that $Z_1 \cup Z_2$ has funny intersection with $Z_3$.  

Ok, let me now give a proof of a correct statement showing that sometimes they are equal.

**Lemma:**  Suppose that subschemes $Z_1, \dots, Z_k$ have pairwise trivial intersection in some ambient Noetherian scheme $X$.  Then $I_{Z_1} \dots I_{Z_K} = I_{Z_1} \cap I_{Z_k}$.  

*Proof:*  The statement is local so we may assume that $X$ is the spectrum of a local ring $(R, \mathfrak{m})$.  Now, since $I_{Z_1} + I_{Z_2} = R$, at least one of those ideals must equal $R$ (if not, both would be in the maximal ideal $\mathfrak{m}$, and so would their sum).  Likewise with all pairs.  Therefore, at most one of the ideals $I_{Z_i}$ is not equal to $R$.  But now the statement is obvious.  $R \cdot R \dots I_{Z_i} \dots R = I_{Z_i} = R \cap R \cap \dots I_{Z_i} \cap \dots R$.