How can I quickly guess if L is context-free or det. context-free?
This answer was first written before the OP edited the question to specify
that he wanted "a quick test I can apply by inspection, without use of
a computer, that will often be correct."

Of course, I do not think anyone has such a test. To begin with, we
should first define a meta-language defining sets of strings to which
such a test could be applied. How general should this language be, as
we can even define metalanguages that can describe only context-free sets.
But that is no help, since the difficulty remains in the translation
between metalanguages.  In other words, this is simply an ill defined
question as asked.

And even if we were to agree on the language, as something that looks
more or less like the examples, we just do not have such
tests. Actually the question was raised before, as a suggestion for
research, but I do not recall there was any follow-up.

This discussion is mostly intended for the determination of
CF-ness. Raphael rightly commented that if the language is known to be
CF, even assuming that a CF grammar is known for it, it remains
undecidable to determine whether it is deterministic CF. However, the
OP is not asking for a decision procedure, but only a test which works
most of the time. So assuming there is a proper definition of what
most of the time means, such a test could still exist, though I do not
know of any. To take an example in a somewhat different family of
problems, there are primality tests that are fast and known to work
most of the time, with a probability of error that can be chosen
arbitrarily low (with increased cost).

Before I knew the OP wanted "a quick test", I thought he was only
asking for the heuristics we use to get a feeling that some language
is more likely to be this or that. I tried to give some hints on how I
do it. There is nothing more to my answer. It is based on a few
patterns that we see often enough, though the devil in the details may
sometimes change the answer.

Here is what I first answered

Since your question is whether we can see things at a glance, the
answer is that for many exercises, we can ... though we may
occasionally make mistakes.

Very often the stack is used simply as a counter, which is only a
small part of the CF power.

This is typically the case of the following example: $\{a^l b^m \mid l, m
\in \mathbb{N}, l=4m\}$, you have to check that 2 subparts of the
string have the same length $m$, up to a multiplicative rational
constant (4 in this case). That is easily done with a pushdown stack,
by pushing $m$ times in the stack the first time, and then popping $m$
times, checking that there is enough and not too much to pop. When you
push, or pop, you can do it for several symbols to account for the
constant.

However, this allows you to synchronize the sizes of two parts of the
input string, but not more. Once the information is popped, it is lost
and there is no way you can check a third part for the same size.
This is the case of your first example, where you would have to check
for $2n$ and twice for $n$, which makes three times, one too many.

The second example can be discounted as non CF for another reason. The
pumping lemma, without going into details, does tell you that an
infinite CF language has strings that can increase in size by constant
increments. That is, there is some number $n$ such that you can find an
infinite sequence of strings that increase in length by exactly $n$.

For a language of the form $\{a^l b^m \mid l, m \in \mathbb{N},
l=4^m\}$, when you increase $m$ by 1, the number of $a$ increases by
$4^{m+1}-4^m=3\times 4^m$.  That is incompatible with the presence of a linearly
increasing sequence. Exponentials usually do that.

This is an example of using the pumping lemma without even bothering
to try pumping anything. Geological considerations tell you the well will be dry.

Hence I conclude that none of your examples is CF,

But that is not always a full proof (though the pumping lemma use could be considered one). Mostly first evaluation at a glance, heuristics. One might combine exponentials so that things work.

As for deterministic, it is another kind of game. Basically, you try
to see whether you have to guess sometning in order to do your
counting right, so that it succeeds.

In the first example I gave, you obviously do not need any guess. It is
deterministic CF.

But if you consider the language
 $\{a^n b^{2n}c \mid n\in \mathbb{N}\} \cup\{a^n b^{4n}d \mid n\in
 \mathbb{N}\}$
it is non deterministic because you have to guess the last symbol of
the string before you decide to count with the constant 2 or with the
constant 4.

Thanks to  Hendrik Jan for pointing out my misreading of one example.