Breaking RSA using Chinese Remainder Theorem When the same message is encrypted for three people who happen to have same public key but different values of n, it is possible to get the value of message by using Chinese Remainder Theorem.

Wherever I have read about this technique, it says that the message value should be smaller than individual values of n. Will someone please elaborate why this restriction? And why does this technique not work if message is greater than any value of n?
I guess you are refering to the case where three people have the following public keys : $(n_1, 3)$, $(n_2, 3)$ and $(n_3, 3)$ (yes, $e = 3$, and this attack is one of the reasons we don't use $e = 3$ anymore).

In this case, if the same message $m$ is sent to those three persons, we have the following system :
$$
\left\{
    \begin{array}{ll}
        c_1 = &amp; m^3\mod n_1 \\
        c_2 = &amp; m^3\mod n_2 \\
        c_3 = &amp; m^3\mod n_3 \\
    \end{array}
\right.
$$

Applying the CRT to
$$
\left\{
    \begin{array}{ll}
        c_1 = &amp; x\mod n_1 \\
        c_2 = &amp; x\mod n_2 \\
        c_3 = &amp; x\mod n_3 \\
    \end{array}
\right.
$$
with $x = m^3$ will give you $x = m^3\mod n_1\times n_2\times n_3$

However, we know that $m \lt n_1,\ n_2,\ n_3$ so we have $ m^3 \lt n_1\times n_2\times n_3$ so a simple cubic root will give us the original message. 

If the message is greater than any $n_i$, you wouldn't be able to recover it with a simple cubic root because you would have a modular result.

Today, we usually use $e=2^{16} + 1 = 65537$ which make this attack much more difficult to apply.