(This is not really an answer, but some formalistic remarks, which got too large for a comment.)
$\def\TDES{\operatorname{3DES}}\def\ECB{\operatorname{ECB}}\def\AES{\operatorname{AES}}\def\Serp{\operatorname{Serp}}\def\CTR{\operatorname{CTR}}$
First, to combine block ciphers (i.e. form a new block cipher from the individual ones), the components block ciphers need to have a common block size - and 3DES has a block size of 64 instead of 128 for the other ones. So we would first have to somehow build a 128-bit block cipher from the 64-bit one, for example using something like 2-block-ECB mode (I'll name this $\TDES'$ in the following).
Now of course $\AES \circ \Serp \circ \TDES'$ is a different block cipher than $\TDES' \circ \AES \circ \Serp$ etc., and the question would be if some of them are harder or easier to break (as a block cipher, used within any mode, or in the standard models.)
On the other hand, if we don't combine the block-cipher block-wise, but together with their mode of operation, it will also depend on the mode of operation.
For modes like CTR and OFB, which actually produce a key stream, which will then be XOR-ed with the data, the answer to your question is:
The order doesn't matter at all, as all variants give the same ciphertext (due to the commutativity and associativity of XOR):
$$ \begin{align*} &\quad \CTR[\AES]\circ \CTR[\TDES]\circ \CTR[\Serp] \\&= \CTR[\Serp] \circ \CTR[\AES]\circ \CTR[\TDES] \\ & = \CTR[\TDES] \circ \CTR[\Serp] \circ \CTR[\AES]\end{align*} $$
(or all other combinations, assuming you are using the same key, mode and initialization vector for the same cipher in each case.) The same is valid for OFB and even mixed uses, as well as for stream ciphers like RC4.
For ECB-mode, we have
$$\begin{align*} \ECB[\AES\circ\Serp \circ \TDES'] &= \ECB[\AES]\circ\ECB[\Serp] \circ \ECB[\TDES'] \\ &= \ECB[\AES]\circ\ECB[\Serp] \circ \ECB[\TDES], \end{align*}$$
(and similar for the other combinations), but you shouldn't use ECB mode anyways.
For CBC and CFB, it gets complicated, though.