Why are condensation reactions endergonic?
One of the most simple biosynthetic reactions that generates ‘oligomers’ from ‘monomers’ (between inverted commas because that description is not fully correct here) is the ester formation used to synthesise lipids. It can be broken down to:

$$\ce{R-O-H + H'-O'-C(=O)-R' -&gt; R-O-C(=O)-R' + H'-O'-H}$$

With each important bond highlighted and the oxygen and hydrogen atoms made distinguishable by primes. A simplistic way of deriving standard reaction enthalpies is to use Hess’ law and add up the bond dissociation enthalpies of bonds that are broken and subtract that from the bond dissociation enthalpies of bonds that are formed. Doing that in this case gives:


Broken:


$\ce{C-O'}$
$\ce{O-H}$

Formed:


$\ce{C-O}$
$\ce{O'-H}$



Since the bonds being formed and those being broken are identical, at first approximation $\Delta_\mathrm{r} H^0 \approx 0$, meaning the reaction depends entirely on the entropic term.

On the entropic side of things, you are creating one huge, dissolved entity out of two not-as-huge dissolved entities. Making an exact call here is difficult, as it depends on a number of things, most notably how large the hydration sphere is for all three of these compounds. If each of them has a hydration sphere of approximately the same diameter, the product side will have more entropy since less water molecules are needed for an identical hydration sphere (akin to the hydrophobic effect). The opposite effect is also possible, though.

I would like to challenge the ‘all condensations are …’ It seems to simplistic to me. Many condensation reactions are probably endergonic but likewise I expect a few to be overall exergonic. The cell just takes no chances and uses an $\ce{ATP}$-dependent pathway in all cases to ensure the outcome it desires.